Added(A)/Deleted(D) following books

 A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter25_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter26_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter27_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter28_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter29_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter30_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter31_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter32_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter33_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter34_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter35_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter36_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter37_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter38_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/chapter39_6.ipynb
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/screenshots/chapter29example32_6.png
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/screenshots/chapter29example33_6.png
A  A_Textbook_of_Electrical_Technology_:_AC_and_DC_Machines_(Volume_-_2)_by_A_K_Theraja_B_L_Thereja/screenshots/chapter32example30_6.png
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter10_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter1_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter2_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter3_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter4_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter5_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter6_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter7_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter8_1.ipynb
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/InverseMatrix_1.png
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/determinant_1.png
A  Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/scalorpolynomial_1.png

12 files changed, 4330 insertions, 0 deletions

diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter10_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter10_1.ipynb
new file mode 100644
index 00000000..1c137a08
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter10_1.ipynb
@@ -0,0 +1,128 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 10 - Bilinear forms"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 363 Example 10.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a = [x1 x2]\n",
+      "b = [y1 y2]\n",
+      "f(a,b) = x1*y1 + x1*y2 + x2*y1 + x2*y2\n",
+      "so, f(a,b) = \n",
+      "[x1  x2]  *  |1   1|  *   |y1|\n",
+      "              |1   1|     |y2|\n",
+      "So the matrix of f in standard order basis B = {e1,e2} is:\n",
+      "[f]B = \n",
+      "[[1 1]\n",
+      " [1 1]]\n",
+      "P = \n",
+      "[[ 1  1]\n",
+      " [-1  1]]\n",
+      "Thus, [f]B = P*[f]B*P\n",
+      "[f]B = \n",
+      "[[ 1 -1]\n",
+      " [-1  1]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "print 'a = [x1 x2]'\n",
+    "print 'b = [y1 y2]'\n",
+    "print 'f(a,b) = x1*y1 + x1*y2 + x2*y1 + x2*y2'\n",
+    "print 'so, f(a,b) = '\n",
+    "print '[x1  x2]  *  |1   1|  *   |y1|'\n",
+    "print '              |1   1|     |y2|'\n",
+    "print 'So the matrix of f in standard order basis B = {e1,e2} is:'\n",
+    "fb = np.array([[1, 1],[1, 1]])\n",
+    "print '[f]B = \\n',fb\n",
+    "P = np.array([[1 ,1],[-1, 1]])\n",
+    "print 'P = \\n',P\n",
+    "print 'Thus, [f]B'' = P''*[f]B*P'\n",
+    "fb1 = np.transpose(P) * fb * P\n",
+    "print '[f]B'' = \\n',fb1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 365 Example 10.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "n =  56.0\n",
+      "a =  410.0\n",
+      "b =  70.0\n",
+      "f(a,b) =  28700.0\n",
+      "f is non-degenerate billinear form on R**n.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "n = round(np.random.randint(2,90))\n",
+    "a = round(np.random.randint(1,n) * 10)#\n",
+    "b = round(np.random.randint(1,n) * 10)#\n",
+    "print 'n = ',n\n",
+    "print 'a = ',a\n",
+    "print 'b = ',b\n",
+    "f = a * np.transpose(b)\n",
+    "print 'f(a,b) = ',f\n",
+    "print 'f is non-degenerate billinear form on R**n.'\n",
+    "#end"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter1_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter1_1.ipynb
new file mode 100644
index 00000000..b45b10cf
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter1_1.ipynb
@@ -0,0 +1,754 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 - Linear Equations"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 8 Example 1.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=\n",
+      "[[ 2 -1  3  2]\n",
+      " [ 1  4  0 -1]\n",
+      " [ 2  6 -1  5]]\n",
+      "Applying row transformations:\n",
+      "R1 = R1-2*R2\n",
+      "a = \n",
+      "[[ 0 -9  3  4]\n",
+      " [ 1  4  0 -1]\n",
+      " [ 2  6 -1  5]]\n",
+      "R3 = R3-2*R2\n",
+      "a = \n",
+      "[[ 0 -9  3  4]\n",
+      " [ 1  4  0 -1]\n",
+      " [ 0 -2 -1  7]]\n",
+      "R3 = R3/-2\n",
+      "a = \n",
+      "[[ 0 -9  3  4]\n",
+      " [ 1  4  0 -1]\n",
+      " [ 0  1  0 -3]]\n",
+      "R2 = R2-4*R3\n",
+      "a = \n",
+      "[[ 0 -9  3  4]\n",
+      " [ 1  0  0 11]\n",
+      " [ 0  1  0 -3]]\n",
+      "R1 = R1+9*R3\n",
+      "a = \n",
+      "[[  0   0   3 -23]\n",
+      " [  1   0   0  11]\n",
+      " [  0   1   0  -3]]\n",
+      "R1 = R1*2/15\n",
+      "a = \n",
+      "[[ 0  0  0 -4]\n",
+      " [ 1  0  0 11]\n",
+      " [ 0  1  0 -3]]\n",
+      "R2 = R2+2*R1\n",
+      "a = \n",
+      "[[ 0  0  0 -4]\n",
+      " [ 1  0  0  3]\n",
+      " [ 0  1  0 -3]]\n",
+      "R3 = R3-R1/2\n",
+      "a = \n",
+      "[[ 0  0  0 -4]\n",
+      " [ 1  0  0  3]\n",
+      " [ 0  1  0 -1]]\n",
+      "We get the system of equations as:\n",
+      "2*x1 - x2 + 3*x3 + 2*x4 = 0\n",
+      "x1 + 4*x2 - x4 = 0\n",
+      "2*x1 + 6* x2 - x3 + 5*x4 = 0\n",
+      "and\n",
+      "x2 - 5/3*x4 = 0 x1 + 17/3*x4 = 0 x3 - 11/3*x4 = 0\n",
+      "now by assigning any rational value c to x4 in system second, the solution is evaluated as:\n",
+      "(-17/3*c,5/3,11/3*c,c)\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a = np.array([[2, -1, 3, 2],[1, 4, 0 ,-1],[2, 6, -1, 5]])\n",
+    "print 'a=\\n',a\n",
+    "print 'Applying row transformations:'\n",
+    "print 'R1 = R1-2*R2'\n",
+    "a[0,:] = a[0,:] - 2*a[1,:]#\n",
+    "print 'a = \\n',a\n",
+    "print 'R3 = R3-2*R2'\n",
+    "a[2,:] = a[2,:] - 2*a[1,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R3 = R3/-2'\n",
+    "a[2,:] = -1.0/2*a[2,:]#\n",
+    "print 'a = \\n',a\n",
+    "print 'R2 = R2-4*R3'\n",
+    "a[1,:] = a[1,:] - 4*a[2,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R1 = R1+9*R3'\n",
+    "a[0,:] = a[0,:] + 9*a[2,:]#\n",
+    "print 'a = \\n',a\n",
+    "print 'R1 = R1*2/15'\n",
+    "a[0,:] = a[0,:] * 2/15\n",
+    "print 'a = \\n',a\n",
+    "print 'R2 = R2+2*R1'\n",
+    "a[1,:] = a[1,:] + 2*a[0,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R3 = R3-R1/2'\n",
+    "a[2,:] = a[2,:] - 1.0/2*a[0,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'We get the system of equations as:'\n",
+    "print '2*x1 - x2 + 3*x3 + 2*x4 = 0'\n",
+    "print 'x1 + 4*x2 - x4 = 0'\n",
+    "print '2*x1 + 6* x2 - x3 + 5*x4 = 0'\n",
+    "print 'and'\n",
+    "print 'x2 - 5/3*x4 = 0','x1 + 17/3*x4 = 0','x3 - 11/3*x4 = 0'\n",
+    "print 'now by assigning any rational value c to x4 in system second, the solution is evaluated as:'\n",
+    "print '(-17/3*c,5/3,11/3*c,c)'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 9 Example 1.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a = \n",
+      "[[-1.+0.j  0.+1.j]\n",
+      " [ 0.-1.j  3.+0.j]\n",
+      " [ 1.+0.j  2.+0.j]]\n",
+      "Applying row transformations:\n",
+      "R1 = R1+R3 and R2 = R2 + i *R3\n",
+      "a = \n",
+      "[[ 0.+0.j  2.+1.j]\n",
+      " [ 0.+0.j  3.+2.j]\n",
+      " [ 1.+0.j  2.+0.j]]\n",
+      "R1 = R1 * (1/2+i)\n",
+      "a = \n",
+      "[[ 0.+0.j  1.+0.j]\n",
+      " [ 0.+0.j  3.+2.j]\n",
+      " [ 1.+0.j  2.+0.j]]\n",
+      "R2 = R2-R1*(3+2i) and R3 = R3 - 2 *R1\n",
+      "a = \n",
+      "[[ 0.+0.j  1.+0.j]\n",
+      " [ 0.+0.j  0.+0.j]\n",
+      " [ 1.+0.j  0.+0.j]]\n",
+      "Thus the system of equations is:\n",
+      "x1 + 2*x2 = 0 -i*x1 + 3*x2 = 0 -x1+i*x2 = 0\n",
+      "It has only trivial solution x1 = x2 = 0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a=np.array([[-1, 1J],[-1J, 3],[1 ,2]])\n",
+    "print 'a = \\n',a\n",
+    "print 'Applying row transformations:'\n",
+    "print 'R1 = R1+R3 and R2 = R2 + i *R3'\n",
+    "a[0,:] = a[0,:] +a[2,:]\n",
+    "a[1,:] = a[1,:] + 1J * a[2,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R1 = R1 * (1/2+i)'\n",
+    "a[0,:] = 1.0/(2 + 1J) * a[0,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R2 = R2-R1*(3+2i) and R3 = R3 - 2 *R1'\n",
+    "a[1,:] = (a[1,:] - (3 + 2 * 1J) * a[0,:])\n",
+    "a[2,:] = (a[2,:] - 2 * a[0,:])\n",
+    "print 'a = \\n',a\n",
+    "print 'Thus the system of equations is:'\n",
+    "print 'x1 + 2*x2 = 0','-i*x1 + 3*x2 = 0','-x1+i*x2 = 0'\n",
+    "print 'It has only trivial solution x1 = x2 = 0'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 9 Example 1.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[ 1.  0.  0.  0.  0.  0.  0.  0.]\n",
+      " [ 0.  1.  0.  0.  0.  0.  0.  0.]\n",
+      " [ 0.  0.  1.  0.  0.  0.  0.  0.]\n",
+      " [ 0.  0.  0.  1.  0.  0.  0.  0.]\n",
+      " [ 0.  0.  0.  0.  1.  0.  0.  0.]\n",
+      " [ 0.  0.  0.  0.  0.  1.  0.  0.]\n",
+      " [ 0.  0.  0.  0.  0.  0.  1.  0.]\n",
+      " [ 0.  0.  0.  0.  0.  0.  0.  1.]]\n",
+      "This is an Identity matrix of order 8 * 8\n",
+      "And It is a row reduced matrix.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "n = np.random.randint(9)\n",
+    "print np.identity(n)\n",
+    "print 'This is an Identity matrix of order %d * %d'%(n,n)\n",
+    "print 'And It is a row reduced matrix.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 12 Example 1.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[[ 1.  0.  0.  0.  0.  0.]\n",
+      " [ 0.  1.  0.  0.  0.  0.]\n",
+      " [ 0.  0.  1.  0.  0.  0.]\n",
+      " [ 0.  0.  0.  1.  0.  0.]\n",
+      " [ 0.  0.  0.  0.  1.  0.]\n",
+      " [ 0.  0.  0.  0.  0.  1.]]\n",
+      "This is an Identity matrix of order 6 * 6\n",
+      "And It is a row reduced matrix.\n",
+      "[[ 0.  0.  0.  0.  0.]]\n",
+      "This is an Zero matrix of order 1 * 5\n",
+      "And It is also a row reduced matrix.\n",
+      "a = \n",
+      "[[ 0.   1.  -3.   0.   0.5]\n",
+      " [ 0.   0.   0.   1.   2. ]\n",
+      " [ 0.   0.   0.   0.   0. ]]\n",
+      "This is a non-trivial row reduced matrix.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "n = np.random.randint(9)\n",
+    "print np.identity(n)\n",
+    "print 'This is an Identity matrix of order %d * %d'%(n,n)\n",
+    "print 'And It is a row reduced matrix.'\n",
+    "m = np.random.randint(0,9)\n",
+    "n = np.random.randint(9)\n",
+    "print np.zeros([m,n])\n",
+    "print 'This is an Zero matrix of order %d * %d'%(m,n)\n",
+    "print 'And It is also a row reduced matrix.'\n",
+    "a = np.array([[0, 1, -3, 0, 1.0/2],[0, 0, 0, 1, 2],[0, 0 ,0 ,0 ,0]])\n",
+    "print 'a = \\n',a\n",
+    "print 'This is a non-trivial row reduced matrix.'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 14 Example 1.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[ 1 -2  1]\n",
+      " [ 2  1  1]\n",
+      " [ 0  5 -1]]\n",
+      "Applying row transformations:\n",
+      "R2 = R2 - 2*R1\n",
+      "A = \n",
+      "[[ 1 -2  1]\n",
+      " [ 0  5 -1]\n",
+      " [ 0  5 -1]]\n",
+      "R3 = R3 - R2\n",
+      "A = \n",
+      "[[ 1 -2  1]\n",
+      " [ 0  5 -1]\n",
+      " [ 0  0  0]]\n",
+      "R2 = 1/5*R2\n",
+      "A = \n",
+      "[[ 1 -2  1]\n",
+      " [ 0  1  0]\n",
+      " [ 0  0  0]]\n",
+      "R1 = R1 - 2*R2\n",
+      "A = \n",
+      "[[1 0 1]\n",
+      " [0 1 0]\n",
+      " [0 0 0]]\n",
+      "The condition that the system have a solution is:\n",
+      "2*y1 - y2 + y3 = 0\n",
+      "where, y1,y2,y3 are some scalars\n",
+      "If the condition is satisfied then solutions are obtained by assigning a value c to x3\n",
+      "Solutions are:\n",
+      "x2 = 1/5*c + 1/5*(y2 - 2*y1) x1 = -3/5*c + 1/5*(y1 + 2*y2)\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "A = np.array([[1, -2, 1],[2, 1, 1],[0, 5, -1]])\n",
+    "print 'A = \\n',A\n",
+    "print 'Applying row transformations:'\n",
+    "print 'R2 = R2 - 2*R1'\n",
+    "A[1,:] = A[1,:] - 2*A[0,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'R3 = R3 - R2'\n",
+    "A[2,:] = A[2,:] - A[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'R2 = 1/5*R2'\n",
+    "A[1,:] = 1.0/5*A[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'R1 = R1 - 2*R2'\n",
+    "A[0,:] = A[0,:] + 2*A[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'The condition that the system have a solution is:'\n",
+    "print '2*y1 - y2 + y3 = 0'\n",
+    "print 'where, y1,y2,y3 are some scalars'\n",
+    "print 'If the condition is satisfied then solutions are obtained by assigning a value c to x3'\n",
+    "print 'Solutions are:'\n",
+    "print 'x2 = 1/5*c + 1/5*(y2 - 2*y1)','x1 = -3/5*c + 1/5*(y1 + 2*y2)'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 17 Example 1.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=\n",
+      "[[ 1  0]\n",
+      " [-3  1]]\n",
+      "b=\n",
+      "[[ 5 -1  2]\n",
+      " [15  4  8]]\n",
+      "ab = \n",
+      "[[ 5 -1  2]\n",
+      " [ 0  7  2]]\n",
+      "-----------------------------------------------------------------\n",
+      "a=\n",
+      "[[ 1  0]\n",
+      " [-2  3]\n",
+      " [ 5  4]\n",
+      " [ 0  1]]\n",
+      "b=\n",
+      "[[ 0  6  1]\n",
+      " [ 3  8 -2]]\n",
+      "ab = \n",
+      "[[ 0  6  1]\n",
+      " [ 9 12 -8]\n",
+      " [12 62 -3]\n",
+      " [ 3  8 -2]]\n",
+      "-----------------------------------------------------------------\n",
+      "a=\n",
+      "[[2 1]\n",
+      " [5 4]]\n",
+      "b=\n",
+      "[[1]\n",
+      " [6]]\n",
+      "ab = \n",
+      "[[ 8]\n",
+      " [29]]\n",
+      "-----------------------------------------------------------------\n",
+      "a=\n",
+      "[[-1]\n",
+      " [ 3]]\n",
+      "b=\n",
+      "[[2 4]]\n",
+      "ab = \n",
+      "[[-2 -4]\n",
+      " [ 6 12]]\n",
+      "-----------------------------------------------------------------\n",
+      "a=\n",
+      "[[2 4]]\n",
+      "b=\n",
+      "[[-1]\n",
+      " [ 3]]\n",
+      "ab = \n",
+      "[[10]]\n",
+      "-----------------------------------------------------------------\n",
+      "a=\n",
+      "[[0 1 0]\n",
+      " [0 0 0]\n",
+      " [0 0 0]]\n",
+      "b=\n",
+      "[[ 1 -5  2]\n",
+      " [ 2  3  4]\n",
+      " [ 9 -1  3]]\n",
+      "ab = \n",
+      "[[2 3 4]\n",
+      " [0 0 0]\n",
+      " [0 0 0]]\n",
+      "-----------------------------------------------------------------\n",
+      "a=\n",
+      "[[ 1 -5  2]\n",
+      " [ 2  3  4]\n",
+      " [ 9 -1  3]]\n",
+      "b=\n",
+      "[[0 1 0]\n",
+      " [0 0 0]\n",
+      " [0 0 0]]\n",
+      "ab = \n",
+      "[[0 1 0]\n",
+      " [0 2 0]\n",
+      " [0 9 0]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "#Part a\n",
+    "a = np.array([[1, 0],[-3, 1]])\n",
+    "b = np.array([[5, -1, 2],[15, 4, 8]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)\n",
+    "\n",
+    "print '-----------------------------------------------------------------'\n",
+    "#Part b\n",
+    "a = np.array([[1, 0],[-2, 3],[5 ,4],[0, 1]])\n",
+    "b = np.array([[0, 6, 1],[3 ,8 ,-2]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)\n",
+    "print '-----------------------------------------------------------------'\n",
+    "#Part c\n",
+    "a = np.array([[2, 1],[5, 4]])\n",
+    "b = np.array([[1],[6]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)\n",
+    "print '-----------------------------------------------------------------'\n",
+    "#Part d\n",
+    "a = np.array([[-1],[3]])\n",
+    "b = np.array([[2, 4]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)\n",
+    "print '-----------------------------------------------------------------'\n",
+    "#Part e\n",
+    "a = np.array([[2, 4]])\n",
+    "b = np.array([[-1],[3]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)\n",
+    "print '-----------------------------------------------------------------'\n",
+    "#Part f\n",
+    "a = np.array([[0, 1 ,0],[0, 0, 0],[0, 0, 0]])\n",
+    "b = np.array([[1, -5, 2],[2, 3, 4],[9 ,-1, 3]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)\n",
+    "print '-----------------------------------------------------------------'\n",
+    "#Part g\n",
+    "a = np.array([[1, -5, 2],[2, 3, 4],[9, -1, 3]])\n",
+    "b = np.array([[0, 1, 0],[0 ,0 ,0],[0, 0, 0]])\n",
+    "print 'a=\\n',a\n",
+    "print 'b=\\n',b\n",
+    "print 'ab = \\n',a.dot(b)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 22 Example 1.14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a = \n",
+      "[[0 1]\n",
+      " [1 0]]\n",
+      "inverse a = \n",
+      "[[ 0.  1.]\n",
+      " [ 1.  0.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a = np.array([[0, 1],[1, 0]])\n",
+    "print 'a = \\n',a\n",
+    "print 'inverse a = \\n',np.linalg.inv(a)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 25 Example 1.15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a = \n",
+      "[[ 2 -1]\n",
+      " [ 1  3]]\n",
+      "Applying row tranformations\n",
+      "Interchange R1 and R2\n",
+      "a = \n",
+      "[[ 1  3]\n",
+      " [ 2 -1]]\n",
+      "R2 = R2 - 2 * R1\n",
+      "a =\n",
+      "  [[ 1  3]\n",
+      " [ 0 -7]]\n",
+      "R2 = R2 *1/(-7)\n",
+      "a = \n",
+      "[[1 3]\n",
+      " [0 1]]\n",
+      "R1 = R1 - 3 * R2\n",
+      "a = \n",
+      "[[1 0]\n",
+      " [0 1]]\n",
+      "Since a  has become an identity matrix. So, a is invertible\n",
+      "inverse of a = \n",
+      "[[ 0.42857143  0.14285714]\n",
+      " [-0.14285714  0.28571429]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a = np.array([[2, -1],[1 ,3]])\n",
+    "b = np.array([[2, -1],[1 ,3]]) #Temporary variable to store a\n",
+    "print 'a = \\n',a\n",
+    "print 'Applying row tranformations'\n",
+    "print 'Interchange R1 and R2'\n",
+    "a[0,:] = a[1,:]\n",
+    "a[1,:] = b[0,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R2 = R2 - 2 * R1'\n",
+    "a[1,:] = a[1,:] - 2 * a[0,:]\n",
+    "print 'a =\\n ',a\n",
+    "print 'R2 = R2 *1/(-7)'\n",
+    "a[1,:] = (-1.0/7) * a[1,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'R1 = R1 - 3 * R2'\n",
+    "a[0,:] = a[0,:] - 3 * a[1,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'Since a  has become an identity matrix. So, a is invertible'\n",
+    "print 'inverse of a = '\n",
+    "print np.linalg.inv(b)# #a was stored in b"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 25 Example 1.16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a = \n",
+      "[[ 1.          0.5         0.33333333]\n",
+      " [ 0.5         0.33333333  0.25      ]\n",
+      " [ 0.33333333  0.25        0.2       ]]\n",
+      "b = \n",
+      "[[ 1.  0.  0.]\n",
+      " [ 0.  1.  0.]\n",
+      " [ 0.  0.  1.]]\n",
+      "Applying row transformations on a and b simultaneously,\n",
+      "R2 = R2 - 1/2 * R1 and R3 = R3 - 1/3*R1\n",
+      "a = \n",
+      "[[ 1.          0.5         0.33333333]\n",
+      " [ 0.          0.08333333  0.08333333]\n",
+      " [ 0.          0.08333333  0.08888889]]\n",
+      "b = \n",
+      "[[ 1.          0.          0.        ]\n",
+      " [-0.5         1.          0.        ]\n",
+      " [-0.33333333  0.          1.        ]]\n",
+      "R3 = R3 - R2\n",
+      "a = \n",
+      "[[  1.00000000e+00   5.00000000e-01   3.33333333e-01]\n",
+      " [  0.00000000e+00   8.33333333e-02   8.33333333e-02]\n",
+      " [  0.00000000e+00   2.77555756e-17   5.55555556e-03]]\n",
+      "b = \n",
+      "[[ 1.          0.          0.        ]\n",
+      " [-0.5         1.          0.        ]\n",
+      " [ 0.16666667 -1.          1.        ]]\n",
+      "R2 = R2 * 12 and R3 = R3 * 180\n",
+      "a = \n",
+      "[[  1.00000000e+00   5.00000000e-01   3.33333333e-01]\n",
+      " [  0.00000000e+00   1.00000000e+00   1.00000000e+00]\n",
+      " [  0.00000000e+00   4.99600361e-15   1.00000000e+00]]\n",
+      "b = \n",
+      "[[   1.    0.    0.]\n",
+      " [  -6.   12.    0.]\n",
+      " [  30. -180.  180.]]\n",
+      "R2 = R2 - R3 and R1 = R1 - 1/3*R3\n",
+      "a = \n",
+      "[[  1.00000000e+00   5.00000000e-01  -4.44089210e-16]\n",
+      " [  0.00000000e+00   1.00000000e+00  -1.33226763e-15]\n",
+      " [  0.00000000e+00   4.99600361e-15   1.00000000e+00]]\n",
+      "b = \n",
+      "[[  -9.   60.  -60.]\n",
+      " [ -36.  192. -180.]\n",
+      " [  30. -180.  180.]]\n",
+      "R1 = R1 - 1/2 * R2\n",
+      "a = \n",
+      "[[ 1.  0.  0.]\n",
+      " [ 0.  1. -0.]\n",
+      " [ 0.  0.  1.]]\n",
+      "b = \n",
+      "[[   9.  -36.   30.]\n",
+      " [ -36.  192. -180.]\n",
+      " [  30. -180.  180.]]\n",
+      "Since, a = identity matrix of order 3*3. So, b is inverse of a\n",
+      "inverse(a) = \n",
+      "[[   9.  -36.   30.]\n",
+      " [ -36.  192. -180.]\n",
+      " [  30. -180.  180.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a = np.array([[1 ,1./2, 1.0/3],[1.0/2 ,1.0/3, 1.0/4],[1.0/3, 1.0/4, 1.0/5]])\n",
+    "print 'a = \\n',a\n",
+    "b = np.identity(3)\n",
+    "print 'b = \\n',b\n",
+    "print 'Applying row transformations on a and b simultaneously,'\n",
+    "print 'R2 = R2 - 1/2 * R1 and R3 = R3 - 1/3*R1'\n",
+    "a[1,:] = a[1,:] - 1.0/2 * a[0,:]\n",
+    "a[2,:] = a[2,:] - 1.0/3 * a[0,:]\n",
+    "b[1,:] = b[1,:] - 1.0/2 * b[0,:]\n",
+    "b[2,:] = b[2,:] - 1.0/3 * b[0,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'b = \\n',b\n",
+    "print 'R3 = R3 - R2'\n",
+    "a[2,:] = a[2,:] -  a[1,:]#\n",
+    "b[2,:] = b[2,:] -  b[1,:]\n",
+    "print 'a = \\n',a\n",
+    "print 'b = \\n',b\n",
+    "print 'R2 = R2 * 12 and R3 = R3 * 180'\n",
+    "a[1,:] = a[1,:] *12#\n",
+    "a[2,:] = a[2,:] * 180#\n",
+    "b[1,:] = b[1,:] * 12#\n",
+    "b[2,:] = b[2,:] * 180#\n",
+    "print 'a = \\n',a\n",
+    "print 'b = \\n',b\n",
+    "print 'R2 = R2 - R3 and R1 = R1 - 1/3*R3'\n",
+    "a[1,:] = a[1,:] - a[2,:]#\n",
+    "a[0,:] = a[0,:] - 1./3 * a[2,:]#\n",
+    "b[1,:] = b[1,:] - b[2,:]#\n",
+    "b[0,:] = b[0,:] - 1./3 * b[2,:]#\n",
+    "print 'a = \\n',a\n",
+    "print 'b = \\n',b\n",
+    "print 'R1 = R1 - 1/2 * R2'\n",
+    "a[0,:] = a[0,:] -  1./2 * a[1,:]#\n",
+    "b[0,:] = b[0,:] -  1./2 * b[1,:]#\n",
+    "print 'a = \\n',np.matrix.round(a)\n",
+    "print 'b = \\n',b\n",
+    "print 'Since, a = identity matrix of order 3*3. So, b is inverse of a'\n",
+    "print 'inverse(a) = \\n',b"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter2_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter2_1.ipynb
new file mode 100644
index 00000000..cac42c12
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter2_1.ipynb
@@ -0,0 +1,822 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 - Vector spaces"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 37 Example 2.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a1 =  [1, 2, 0, 3, 0]\n",
+      "a2 =  [0, 0, 1, 4, 0]\n",
+      "a3 =  [0, 0, 0, 0, 1]\n",
+      "By theorem 3, vector a is in subspace W of F**5 spanned by a1, a2, a3\n",
+      "if and only if there exist scalars c1, c2, c3 such that\n",
+      "a= c1a1 + c2a2 + c3a3\n",
+      "So, a = (c1,2*c1,c2,3c1+4c2,c3)\n",
+      "c1 =  -3\n",
+      "c2 =  1\n",
+      "c3 =  2\n",
+      "Therefore, a =  [0, 0, 1, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1]\n",
+      "This shows, a is in W\n",
+      "And (2,4,6,7,8) is not in W as there is no value of c1 c2 c3 that satisfies the equation\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "a1 = [1, 2 ,0 ,3, 0]#\n",
+    "a2 =[0, 0 ,1 ,4 ,0]#\n",
+    "a3 = [0 ,0 ,0 ,0, 1]#\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print 'By theorem 3, vector a is in subspace W of F**5 spanned by a1, a2, a3'\n",
+    "print 'if and only if there exist scalars c1, c2, c3 such that'\n",
+    "print 'a= c1a1 + c2a2 + c3a3'\n",
+    "print 'So, a = (c1,2*c1,c2,3c1+4c2,c3)'\n",
+    "c1 = -3#\n",
+    "c2 = 1#\n",
+    "c3 = 2#\n",
+    "a = c1*a1 + c2*a2 + c3*a3#\n",
+    "print 'c1 = ',c1\n",
+    "print 'c2 = ',c2\n",
+    "print 'c3 = ',c3\n",
+    "print 'Therefore, a = ',a\n",
+    "print 'This shows, a is in W'\n",
+    "print 'And (2,4,6,7,8) is not in W as there is no value of c1 c2 c3 that satisfies the equation'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 38 Example 2.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[1 2 0 3 0]\n",
+      " [0 0 1 4 0]\n",
+      " [0 0 0 0 1]]\n",
+      "The subspace of F**5 spanned by a1 a2 a3(row vectors of A) is called row space of A.\n",
+      "a1 =  [1 2 0 3 0]\n",
+      "a2 =  [0 0 1 4 0]\n",
+      "a3 =  [0 0 0 0 1]\n",
+      "And, it is also the row space of B.\n",
+      "B = \n",
+      "[[ 1  2  0  3  0]\n",
+      " [ 0  0  1  4  0]\n",
+      " [ 0  0  0  0  1]\n",
+      " [-4 -8  1 -8  0]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "A = np.array([[1, 2, 0 ,3 ,0],[0, 0, 1, 4, 0],[0, 0, 0, 0, 1]])\n",
+    "print 'A = \\n',A\n",
+    "print 'The subspace of F**5 spanned by a1 a2 a3(row vectors of A) is called row space of A.'\n",
+    "a1 = A[0,:]\n",
+    "a2 = A[1,:]\n",
+    "a3 = A[2,:]\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print 'And, it is also the row space of B.'\n",
+    "B = np.array([[1, 2, 0, 3, 0],[0, 0, 1, 4, 0],[0, 0, 0, 0, 1],[-4, -8, 1 ,-8, 0]])\n",
+    "print 'B = \\n',B"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 39 Example 2.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "V is the space of all polynomial functions over F.\n",
+      "S contains the functions as:\n",
+      "n =  17\n",
+      "f0(x) =  1\n",
+      "f1(x) =  x\n",
+      "f2(x) =  x**2\n",
+      "f3(x) =  x**3\n",
+      "f4(x) =  x**4\n",
+      "f5(x) =  x**5\n",
+      "f6(x) =  x**6\n",
+      "f7(x) =  x**7\n",
+      "f8(x) =  x**8\n",
+      "f9(x) =  x**9\n",
+      "f10(x) =  x**10\n",
+      "f11(x) =  x**11\n",
+      "f12(x) =  x**12\n",
+      "f13(x) =  x**13\n",
+      "f14(x) =  x**14\n",
+      "f15(x) =  x**15\n",
+      "f16(x) =  x**16\n",
+      "Then, V is the subspace spanned by set S.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "import numpy as np\n",
+    "print 'V is the space of all polynomial functions over F.'\n",
+    "print 'S contains the functions as:'\n",
+    "x = sp.Symbol(\"x\")\n",
+    "#n = round(rand()*10)#\n",
+    "n=np.random.randint(0,19)\n",
+    "print 'n = ',n\n",
+    "for i in range (0,n):\n",
+    "    f = x**i#\n",
+    "    print 'f%d(x) = '%(i,),f\n",
+    "    \n",
+    "\n",
+    "print 'Then, V is the subspace spanned by set S.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 41 Example 2.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a1 =  [ 3  0 -3]\n",
+      "a2 =  [-1  1  2]\n",
+      "a3 =  [ 4  2 -2]\n",
+      "a4 =  [2 1 1]\n",
+      " Since, 2 * a1 + 2 * a2 - a3 + 0 * a4 =  [0 0 0] = 0\n",
+      "a1,a2,a3,a4 are linearly independent\n",
+      "Now, e1 =  [1, 0, 0]\n",
+      "e2 =  [0, 1, 0]\n",
+      "e3 =  [0, 0, 1]\n",
+      "Also, e1,e2,e3 are linearly independent.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a1 = np.array([3 ,0, -3])\n",
+    "a2 = np.array([-1 ,1 ,2])\n",
+    "a3 = np.array([4 ,2, -2])\n",
+    "a4 = np.array([2 ,1, 1])\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print 'a4 = ',a4\n",
+    "t = 2 * a1 + 2 * a2 - a3 + 0 * a4\n",
+    "print ' Since, 2 * a1 + 2 * a2 - a3 + 0 * a4 = ',t,'= 0'\n",
+    "print 'a1,a2,a3,a4 are linearly independent'\n",
+    "e1 = [1, 0, 0]#\n",
+    "e2 = [0 ,1 ,0]#\n",
+    "e3 = [0 ,0, 1]#\n",
+    "print  'Now, e1 = ',e1\n",
+    "print  'e2 = ',e2\n",
+    "print  'e3 = ',e3\n",
+    "print 'Also, e1,e2,e3 are linearly independent.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 41 Example 2.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "S is the subset of F**n consisting of n vectors.\n",
+      "n =  10\n",
+      "e1 = \n",
+      "[ 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.]\n",
+      "e2 = \n",
+      "[ 0.  1.  0.  0.  0.  0.  0.  0.  0.  0.]\n",
+      "e3 = \n",
+      "[ 0.  0.  1.  0.  0.  0.  0.  0.  0.  0.]\n",
+      "e4 = \n",
+      "[ 0.  0.  0.  1.  0.  0.  0.  0.  0.  0.]\n",
+      "e5 = \n",
+      "[ 0.  0.  0.  0.  1.  0.  0.  0.  0.  0.]\n",
+      "e6 = \n",
+      "[ 0.  0.  0.  0.  0.  1.  0.  0.  0.  0.]\n",
+      "e7 = \n",
+      "[ 0.  0.  0.  0.  0.  0.  1.  0.  0.  0.]\n",
+      "e8 = \n",
+      "[ 0.  0.  0.  0.  0.  0.  0.  1.  0.  0.]\n",
+      "e9 = \n",
+      "[ 0.  0.  0.  0.  0.  0.  0.  0.  1.  0.]\n",
+      "e10 = \n",
+      "[ 0.  0.  0.  0.  0.  0.  0.  0.  0.  1.]\n",
+      "x1,x2,x3...xn are the scalars in F\n",
+      "Putting a = x1*e1 + x2*e2 + x3*e3 + .... + xn*en\n",
+      "So, a = (x1,x2,x3,...,xn)\n",
+      "Therefore, e1,e2..,en span F**n\n",
+      "a = 0 if x1 = x2 = x3 = .. = xn = 0\n",
+      "So,e1,e2,e3,..,en are linearly independent.\n",
+      "The set S = {e1,e2,..,en} is called standard basis of F**n\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "print 'S is the subset of F**n consisting of n vectors.'\n",
+    "#n = round(rand() *10 + 1)#\\\n",
+    "n=np.random.randint(0,19)\n",
+    "print 'n = ',n\n",
+    "I = np.identity(n)\n",
+    "for i in range(0,n):\n",
+    "    e = I[i,:]\n",
+    "    print 'e%d = '%(i+1)\n",
+    "    print e\n",
+    "\n",
+    "print 'x1,x2,x3...xn are the scalars in F'\n",
+    "print 'Putting a = x1*e1 + x2*e2 + x3*e3 + .... + xn*en'\n",
+    "print 'So, a = (x1,x2,x3,...,xn)'\n",
+    "print 'Therefore, e1,e2..,en span F**n'\n",
+    "print 'a = 0 if x1 = x2 = x3 = .. = xn = 0'\n",
+    "print 'So,e1,e2,e3,..,en are linearly independent.'\n",
+    "print 'The set S = {e1,e2,..,en} is called standard basis of F**n'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 54 Example 2.20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "P = \n",
+      "[[-1  4  5]\n",
+      " [ 0  2 -3]\n",
+      " [ 0  0  8]]\n",
+      "\n",
+      "inverse(P) = \n",
+      "[[-1.      2.      1.375 ]\n",
+      " [ 0.      0.5     0.1875]\n",
+      " [ 0.      0.      0.125 ]]\n",
+      "The vectors forming basis of F**3 are a1, a2, a3\n",
+      "a1' = \n",
+      "[-1  0  0]\n",
+      "\n",
+      "a2' = \n",
+      "[4 2 0]\n",
+      "\n",
+      "a3' = \n",
+      "[ 5 -3  8]\n",
+      "The coordinates x1,x2,x3 of vector a = [x1,x2,x3] is given by inverse(P)*[x1# x2# x3]\n",
+      "And, -10*a1 - 1/2*a2 - a3 =  [ 3.  2. -8.]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "P = np.array([[-1, 4, 5],[ 0, 2, -3],[ 0, 0, 8]])\n",
+    "print 'P = \\n',P\n",
+    "print '\\ninverse(P) = \\n',np.linalg.inv(P)\n",
+    "a1 = P[:,0]\n",
+    "a2 = P[:,1]\n",
+    "a3 = P[:,2]\n",
+    "print 'The vectors forming basis of F**3 are a1'', a2'', a3'''\n",
+    "print \"a1' = \\n\",np.transpose(a1)\n",
+    "print \"\\na2' = \\n\",np.transpose(a2)\n",
+    "print \"\\na3' = \\n\",np.transpose(a3)\n",
+    "print 'The coordinates x1'',x2'',x3'' of vector a = [x1,x2,x3] is given by inverse(P)*[x1# x2# x3]'\n",
+    "t = -10*a1 - 1./2*a2 - a3#\n",
+    "print 'And, -10*a1'' - 1/2*a2'' - a3'' = ',t"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 60 Example 2.21"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Given row vectors are:\n",
+      "a1 =  [1, 2, 2, 1]\n",
+      "a2 =  [0, 2, 0, 1]\n",
+      "a3 =  [-2, 0, -4, 3]\n",
+      "The matrix A from these vectors will be:\n",
+      "A = \n",
+      "[[ 1  2  2  1]\n",
+      " [ 0  2  0  1]\n",
+      " [-2  0 -4  3]]\n",
+      "Finding Row reduced echelon matrix of A that is given by R\n",
+      "And applying same operations on identity matrix Q such that R = QA\n",
+      "Q = \n",
+      "[[ 1.  0.  0.]\n",
+      " [ 0.  1.  0.]\n",
+      " [ 0.  0.  1.]]\n",
+      "Applying row transformations on A and Q,we get\n",
+      "R1 = R1-R2\n",
+      "A = \n",
+      "[[ 1  0  2  0]\n",
+      " [ 0  2  0  1]\n",
+      " [-2  0 -4  3]]\n",
+      "Q = \n",
+      "[[ 1. -1.  0.]\n",
+      " [ 0.  1.  0.]\n",
+      " [ 0.  0.  1.]]\n",
+      "R3 = R3 + 2*R1\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 2 0 1]\n",
+      " [0 0 0 3]]\n",
+      "Q = \n",
+      "[[ 1. -1.  0.]\n",
+      " [ 0.  1.  0.]\n",
+      " [ 2. -2.  1.]]\n",
+      "R3 = R3/3\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 2 0 1]\n",
+      " [0 0 0 1]]\n",
+      "Q = \n",
+      "[[ 1.         -1.          0.        ]\n",
+      " [ 0.          1.          0.        ]\n",
+      " [ 0.66666667 -0.66666667  0.33333333]]\n",
+      "R2 = R2/2\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 1 0 0]\n",
+      " [0 0 0 1]]\n",
+      "Q = \n",
+      "[[ 1.         -1.          0.        ]\n",
+      " [ 0.          5.          0.        ]\n",
+      " [ 0.66666667 -0.66666667  0.33333333]]\n",
+      "R2 = R2 - 1/2*R3\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 1 0 0]\n",
+      " [0 0 0 1]]\n",
+      "Q = \n",
+      "[[ 1.         -1.          0.        ]\n",
+      " [-0.33333333  5.33333333 -0.16666667]\n",
+      " [ 0.66666667 -0.66666667  0.33333333]]\n",
+      "Row reduced echelon matrix:\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 1 0 0]\n",
+      " [0 0 0 1]]\n",
+      "Q = \n",
+      "[[ 1.         -1.          0.        ]\n",
+      " [-0.33333333  5.33333333 -0.16666667]\n",
+      " [ 0.66666667 -0.66666667  0.33333333]]\n",
+      "rank of R =  2\n",
+      "Since, Rank of R is 3, so a1, a2, a3 are independent\n",
+      "Now, basis for W can be given by row vectors of R i.e. p1,p2,p3\n",
+      "b is any vector in W. b = [b1 b2 b3 b4]\n",
+      "Span of vectors p1,p2,p3 consist of vector b with b3 = 2*b1\n",
+      "So,b = b1p1 + b2p2 + b4p3\n",
+      "And,[p1 p2 p3] = R = Q*A\n",
+      "So, b = [b1 b2 b3]* Q * A\n",
+      "hence, b = x1a1 + x2a2 + x3a3 where x1 = [b1 b2 b4] * Q(1) and so on\n",
+      "Now, given 3 vectors a1 a2 a3:\n",
+      "a1 =  [1, 0, 2, 0]\n",
+      "a2 =  [0, 2, 0, 1]\n",
+      "a3 =  [0, 0, 0, 3]\n",
+      "Since a1 a2 a3 are all of the form (y1 y2 y3 y4) with y3 = 2*y1, hence they are in W.\n",
+      "So, they are independent.\n",
+      "Required matrix P such that X = PX is:\n",
+      "P = \n",
+      "[[ 0. -0.  2.]\n",
+      " [-0.  0. -2.]\n",
+      " [ 0. -0.  1.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a1 = [1 ,2 ,2, 1]#\n",
+    "a2 = [0 ,2 ,0 ,1]#\n",
+    "a3 = [-2, 0, -4, 3]#\n",
+    "print 'Given row vectors are:'\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print 'The matrix A from these vectors will be:'\n",
+    "#A = [a1],[a2], [a3]]\n",
+    "A=np.array([a1,a2,a3])\n",
+    "print 'A = \\n',A\n",
+    "\n",
+    "print 'Finding Row reduced echelon matrix of A that is given by R'\n",
+    "print 'And applying same operations on identity matrix Q such that R = QA'\n",
+    "Q = np.identity(3)\n",
+    "print 'Q = \\n',Q\n",
+    "T = A#              #Temporary matrix to store A\n",
+    "print 'Applying row transformations on A and Q,we get'\n",
+    "print 'R1 = R1-R2'\n",
+    "A[0,:] = A[0,:] - A[1,:]\n",
+    "Q[0,:] = Q[0,:] - Q[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'Q = \\n',Q\n",
+    "print 'R3 = R3 + 2*R1'\n",
+    "A[2,:] = A[2,:] + 2*A[0,:]\n",
+    "Q[2,:] = Q[2,:] + 2*Q[0,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'Q = \\n',Q\n",
+    "print 'R3 = R3/3'\n",
+    "A[2,:] = 1./3*A[2,:]\n",
+    "Q[2,:] = 1./3*Q[2,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'Q = \\n',Q\n",
+    "print 'R2 = R2/2'\n",
+    "A[1,:] = 1./2*A[1,:]\n",
+    "Q[1,:] = 10/2*Q[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'Q = \\n',Q\n",
+    "print 'R2 = R2 - 1/2*R3'\n",
+    "A[1,:] = A[1,:] - 1./2*A[2,:]\n",
+    "Q[1,:] = Q[1,:] - 1./2*Q[2,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'Q = \\n',Q\n",
+    "R = A#\n",
+    "A = T#\n",
+    "print 'Row reduced echelon matrix:'\n",
+    "print 'A = \\n',A\n",
+    "print 'Q = \\n',Q\n",
+    "#part a\n",
+    "print 'rank of R = ',np.rank(R)\n",
+    "\n",
+    "print 'Since, Rank of R is 3, so a1, a2, a3 are independent'\n",
+    "#part b\n",
+    "print 'Now, basis for W can be given by row vectors of R i.e. p1,p2,p3'\n",
+    "print 'b is any vector in W. b = [b1 b2 b3 b4]'\n",
+    "print 'Span of vectors p1,p2,p3 consist of vector b with b3 = 2*b1'\n",
+    "print 'So,b = b1p1 + b2p2 + b4p3'\n",
+    "print 'And,[p1 p2 p3] = R = Q*A'\n",
+    "print 'So, b = [b1 b2 b3]* Q * A'\n",
+    "print 'hence, b = x1a1 + x2a2 + x3a3 where x1 = [b1 b2 b4] * Q(1) and so on' # #Equation 1\n",
+    "#part c\n",
+    "print 'Now, given 3 vectors a1'' a2'' a3'':'\n",
+    "c1 = [1, 0, 2, 0]#\n",
+    "c2 = [0 ,2 ,0, 1]#\n",
+    "c3 = [0 ,0 ,0 ,3]#\n",
+    "print 'a1'' = ',c1\n",
+    "print 'a2'' = ',c2\n",
+    "print 'a3'' = ',c3\n",
+    "print 'Since a1'' a2'' a3'' are all of the form (y1 y2 y3 y4) with y3 = 2*y1, hence they are in W.'\n",
+    "print 'So, they are independent.'\n",
+    "#part d\n",
+    "c = np.array([c1,c2,c3])\n",
+    "P = np.identity(3)\n",
+    "for i in range(0,3):\n",
+    "    b1 = c[i,0]\n",
+    "    b2 = c[i,1]\n",
+    "    b4 = c[i,3]\n",
+    "    x1 = np.array([b1, b2, b4]) * Q[:,0]\n",
+    "    x2 = np.array([b1, b2, b4])*Q[:,1]\n",
+    "    x3 = np.array([b1, b2, b4])*Q[:,2]\n",
+    "    \n",
+    "\n",
+    "print 'Required matrix P such that X = PX'' is:'\n",
+    "P=np.vstack([x1,x2,x3])\n",
+    "print 'P = \\n',P\n",
+    "#print x1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 63 Example 2.22"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[ 1  2  0  3  0]\n",
+      " [ 1  2 -1 -1  0]\n",
+      " [ 0  0  1  4  0]\n",
+      " [ 2  4  1 10  1]\n",
+      " [ 0  0  0  0  1]]\n",
+      "Taking an identity matrix P:\n",
+      "P = \n",
+      "[[ 1.  0.  0.  0.  0.]\n",
+      " [ 0.  1.  0.  0.  0.]\n",
+      " [ 0.  0.  1.  0.  0.]\n",
+      " [ 0.  0.  0.  1.  0.]\n",
+      " [ 0.  0.  0.  0.  1.]]\n",
+      "Applying row transformations on P and A to get a row reduced echelon matrix R:\n",
+      "R2 = R2 - R1 and R4 = R4 - 2* R1\n",
+      "A = \n",
+      "[[ 1  2  0  3  0]\n",
+      " [ 0  0 -1 -4  0]\n",
+      " [ 0  0  1  4  0]\n",
+      " [ 0  0  1  4  1]\n",
+      " [ 0  0  0  0  1]]\n",
+      "P = \n",
+      "[[ 1.  0.  0.  0.  0.]\n",
+      " [-1.  1.  0.  0.  0.]\n",
+      " [ 0.  0.  1.  0.  0.]\n",
+      " [-2.  0.  0.  1.  0.]\n",
+      " [ 0.  0.  0.  0.  1.]]\n",
+      "R2 = -R2 , R3 = R3 - R1 + R2 and R4  = R4 - R1 + R2\n",
+      "A = \n",
+      "[[1 2 0 3 0]\n",
+      " [0 0 1 4 0]\n",
+      " [0 0 0 0 0]\n",
+      " [0 0 0 0 1]\n",
+      " [0 0 0 0 1]]\n",
+      "P = \n",
+      "[[ 1.  0.  0.  0.  0.]\n",
+      " [ 1. -1. -0. -0. -0.]\n",
+      " [-1.  1.  1.  0.  0.]\n",
+      " [-3.  1.  0.  1.  0.]\n",
+      " [-3.  1.  0.  1.  0.]]\n",
+      "Mutually interchanging R3, R4 and R5\n",
+      "Row reduced echelon matrix R = \n",
+      "[[1 2 0 3 0]\n",
+      " [0 0 1 4 0]\n",
+      " [0 0 0 0 1]\n",
+      " [0 0 0 0 1]\n",
+      " [0 0 0 0 0]]\n",
+      "Invertible Matrix P = \n",
+      "[[ 1.  0.  0.  0.  0.]\n",
+      " [ 1. -1. -0. -0. -0.]\n",
+      " [-3.  1.  0.  1.  0.]\n",
+      " [-3.  1.  0.  1.  0.]\n",
+      " [ 0.  0.  0.  0.  0.]]\n",
+      "Invertible matrix P is not unique. There can be many that depends on operations used to reduce A\n",
+      "-----------------------------------------\n",
+      "For the basis of row space W of A, we can take the non-zero rows of R\n",
+      "It can be given by p1, p2, p3\n",
+      "p1 =  [1 2 0 3 0]\n",
+      "p2 =  [0 0 1 4 0]\n",
+      "p3 =  [0 0 0 0 1]\n",
+      "-----------------------------------------\n",
+      "The row space W consists of vectors of the form:\n",
+      "b = c1p1 + c2p2 + c3p3\n",
+      "i.e. b = (c1,2*c1,c2,3*c1+4*c2,c3) where, c1 c2 c3 are scalars.\n",
+      "So, if b2 = 2*b1 and b4 = 3*b1 + 4*b3  =>  (b1,b2,b3,b4,b5) = b1p1 + b3p2 + b5p3\n",
+      "then,(b1,b2,b3,b4,b5) is in W\n",
+      "-----------------------------------------\n",
+      "The coordinate matrix of the vector (b1,2*b1,b2,3*b1+4*b2,b3) in the basis (p1,p2,p3) is column matrix of b1,b2,b3 such that:\n",
+      "  b1\n",
+      "  b2\n",
+      "  b3\n",
+      "-----------------------------------------\n",
+      "Now, to write each vector in W as a linear combination of rows of A:\n",
+      "Let b = (b1,b2,b3,b4,b5) and if b is in W, then\n",
+      "we know,b = (b1,2*b1,b3,3*b1 + 4*b3,b5)  =>  [b1,b3,b5,0,0]*R\n",
+      "=> b = [b1,b3,b5,0,0] * P*A  =>  b = [b1+b3,-b3,0,0,b5] * A\n",
+      "if b = (-5,-10,1,-11,20)\n",
+      "b = ( [-4, -1, 0, 0, 20] ) *[ [[1 2 0 3 0]\n",
+      " [0 0 1 4 0]\n",
+      " [0 0 0 0 1]\n",
+      " [0 0 0 0 1]\n",
+      " [0 0 0 0 0]] ]\n",
+      "-----------------------------------------\n",
+      "The equations in system RX = 0 are given by R * [x1 x2 x3 x4 x5]\n",
+      "i.e., x1 + 2*x2 + 3*x4\n",
+      "x3 + 4*x4\n",
+      "x5\n",
+      "so, V consists of all columns of the form\n",
+      "[ X=\n",
+      "  -2*x2 - 3*x4\n",
+      "  x2\n",
+      "  -4*x4\n",
+      "  x4\n",
+      "  0\n",
+      "where x2 and x4  are arbitrary ]\n",
+      "-----------------------------------------\n",
+      "Let x2 = 1,x4 = 0 then the given column forms a basis of V\n",
+      "[[-2], [1], [0], [0], [0]]\n",
+      "Similarly,if x2 = 0,x4 = 1 then the given column forms a basis of V\n",
+      "[[-3], [0], [-4], [1], [0]]\n",
+      "-----------------------------------------\n",
+      "The equation AX = Y has solutions X if and only if\n",
+      "-y1 + y2 + y3 = 0\n",
+      "-3*y1 + y2 + y4 -y5 = 0\n",
+      "where, Y = (y1 y2 y3 y4 y5)\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "A = np.array([[1, 2, 0, 3, 0],[1, 2, -1, -1, 0],[0 ,0 ,1 ,4 ,0],[2, 4 ,1 ,10, 1],[0 ,0 ,0 ,0 ,1]])\n",
+    "print 'A = \\n',A\n",
+    "#part a\n",
+    "T = A#                  #Temporary storing A in T\n",
+    "print 'Taking an identity matrix P:'\n",
+    "P = np.identity(5)\n",
+    "print 'P = \\n',P\n",
+    "print 'Applying row transformations on P and A to get a row reduced echelon matrix R:'\n",
+    "print 'R2 = R2 - R1 and R4 = R4 - 2* R1'\n",
+    "A[1,:] = A[1,:] - A[0,:]\n",
+    "P[1,:] = P[1,:] - P[0,:]\n",
+    "A[3,:] = A[3,:] - 2 * A[0,:]\n",
+    "P[3,:] = P[3,:] - 2 * P[0,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'P = \\n',P\n",
+    "print 'R2 = -R2 , R3 = R3 - R1 + R2 and R4  = R4 - R1 + R2'\n",
+    "A[1,:] = -A[1,:]\n",
+    "P[1,:] = -P[1,:]\n",
+    "A[2,:] = A[2,:] - A[1,:]\n",
+    "P[2,:] = P[2,:] - P[1,:]\n",
+    "A[3,:] = A[3,:] - A[1,:]\n",
+    "P[3:] = P[3,:] - P[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'P = \\n',P\n",
+    "print 'Mutually interchanging R3, R4 and R5'\n",
+    "x = A[2,:]\n",
+    "A[2,:] = A[4,:]\n",
+    "y = A[3,:]\n",
+    "A[3,:] = x#\n",
+    "A[4,:] = y - A[2,:]\n",
+    "x = P[2,:]\n",
+    "P[2,:] = P[4,:]\n",
+    "y = P[3,:]\n",
+    "P[3,:] = x#\n",
+    "P[4,:] = y - P[2,:]\n",
+    "R = A#\n",
+    "A = T#\n",
+    "print 'Row reduced echelon matrix R = \\n',R\n",
+    "print 'Invertible Matrix P = \\n',P\n",
+    "print 'Invertible matrix P is not unique. There can be many that depends on operations used to reduce A'\n",
+    "print '-----------------------------------------'\n",
+    "#part b\n",
+    "print 'For the basis of row space W of A, we can take the non-zero rows of R'\n",
+    "print 'It can be given by p1, p2, p3'\n",
+    "p1 = R[0,:]\n",
+    "p2 = R[1,:]\n",
+    "p3 = R[2,:]\n",
+    "print 'p1 = ',p1\n",
+    "print 'p2 = ',p2\n",
+    "print 'p3 = ',p3\n",
+    "print '-----------------------------------------'\n",
+    "#part c\n",
+    "print 'The row space W consists of vectors of the form:'\n",
+    "print 'b = c1p1 + c2p2 + c3p3'\n",
+    "print 'i.e. b = (c1,2*c1,c2,3*c1+4*c2,c3) where, c1 c2 c3 are scalars.'\n",
+    "print 'So, if b2 = 2*b1 and b4 = 3*b1 + 4*b3  =>  (b1,b2,b3,b4,b5) = b1p1 + b3p2 + b5p3'\n",
+    "print 'then,(b1,b2,b3,b4,b5) is in W'\n",
+    "print '-----------------------------------------'\n",
+    "#part d\n",
+    "print 'The coordinate matrix of the vector (b1,2*b1,b2,3*b1+4*b2,b3) in the basis (p1,p2,p3) is column matrix of b1,b2,b3 such that:'\n",
+    "print '  b1'\n",
+    "print '  b2'\n",
+    "print '  b3'\n",
+    "print '-----------------------------------------'\n",
+    "#part e\n",
+    "print 'Now, to write each vector in W as a linear combination of rows of A:'\n",
+    "print 'Let b = (b1,b2,b3,b4,b5) and if b is in W, then'\n",
+    "print 'we know,b = (b1,2*b1,b3,3*b1 + 4*b3,b5)  =>  [b1,b3,b5,0,0]*R'\n",
+    "print '=> b = [b1,b3,b5,0,0] * P*A  =>  b = [b1+b3,-b3,0,0,b5] * A'\n",
+    "print 'if b = (-5,-10,1,-11,20)'\n",
+    "b1 = -5#\n",
+    "b2 = -10#\n",
+    "b3 = 1#\n",
+    "b4 = -11#\n",
+    "b5 = 20#\n",
+    "x = [b1 + b3,-b3,0,0,b5]#\n",
+    "print 'b = (',x,')','*[',A,']'\n",
+    "print '-----------------------------------------'\n",
+    "#part f\n",
+    "print 'The equations in system RX = 0 are given by R * [x1 x2 x3 x4 x5]'\n",
+    "print 'i.e., x1 + 2*x2 + 3*x4'\n",
+    "print 'x3 + 4*x4'\n",
+    "print 'x5'\n",
+    "print 'so, V consists of all columns of the form'\n",
+    "print '[','X='\n",
+    "print '  -2*x2 - 3*x4'\n",
+    "print '  x2'\n",
+    "print '  -4*x4'\n",
+    "print '  x4'\n",
+    "print '  0'\n",
+    "print 'where x2 and x4  are arbitrary',']'\n",
+    "print '-----------------------------------------'\n",
+    "#part g\n",
+    "print 'Let x2 = 1,x4 = 0 then the given column forms a basis of V'\n",
+    "\n",
+    "x2 = 1#\n",
+    "x4 = 0#\n",
+    "print [[-2*x2-3*x4],[ x2],[ -4*x4],[ x4],[ 0]]\n",
+    "print 'Similarly,if x2 = 0,x4 = 1 then the given column forms a basis of V'\n",
+    "x2 = 0#\n",
+    "x4 = 1#\n",
+    "print [[-2*x2-3*x4],[ x2],[ -4*x4],[ x4],[ 0]]\n",
+    "print '-----------------------------------------'\n",
+    "#part h\n",
+    "print 'The equation AX = Y has solutions X if and only if'\n",
+    "print '-y1 + y2 + y3 = 0'\n",
+    "print '-3*y1 + y2 + y4 -y5 = 0'\n",
+    "print 'where, Y = (y1 y2 y3 y4 y5)'"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter3_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter3_1.ipynb
new file mode 100644
index 00000000..2e2e4be2
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter3_1.ipynb
@@ -0,0 +1,637 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 3 - Linear transformation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 70 Example 3.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a1 =  [1, 2]\n",
+      "a2 =  [3, 4]\n",
+      "a1 and a2 are linearly independent and hence form a basis for R**2\n",
+      "According to theorem 1, there is a linear transformation from R**2 to R**3 with the transformation functions as:\n",
+      "Ta1 =  [3, 2, 1]\n",
+      "Ta2 =  [6, 5, 4]\n",
+      "Now, we find scalars c1 and c2 for that we know T(c1a1 + c2a2) = c1(Ta1) + c2(Ta2))\n",
+      "if(1,0) = c1(1,2) + c2(3,4), then \n",
+      "c1 =  1\n",
+      "c2 =  3\n",
+      "The transformation function T(1,0) will be:\n",
+      "T(1,0) =  [3, 2, 1, 6, 5, 4, 6, 5, 4, 6, 5, 4]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a1 = [1, 2]#\n",
+    "a2 = [3 ,4]#\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a1 and a2 are linearly independent and hence form a basis for R**2'\n",
+    "print 'According to theorem 1, there is a linear transformation from R**2 to R**3 with the transformation functions as:'\n",
+    "Ta1 = [3 ,2 ,1]#\n",
+    "Ta2 = [6, 5, 4]#\n",
+    "print 'Ta1 = ',Ta1\n",
+    "print 'Ta2 = ',Ta2\n",
+    "print 'Now, we find scalars c1 and c2 for that we know T(c1a1 + c2a2) = c1(Ta1) + c2(Ta2))'\n",
+    "print 'if(1,0) = c1(1,2) + c2(3,4), then '\n",
+    "#c = inv([a1#a2]') * [1#0]#\n",
+    "c=np.array([a1,a2]).dot(np.array([[1],[0]]))\n",
+    "c1 = c[0,0]\n",
+    "c2 = c[1,0]\n",
+    "print 'c1 = ',c1\n",
+    "print 'c2 = ',c2\n",
+    "print 'The transformation function T(1,0) will be:'\n",
+    "T = c1*Ta1 + c2*Ta2#\n",
+    "print 'T(1,0) = ',T\n",
+    "#end"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 81 Example 3.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x1 =  5\n",
+      "x2 =  2\n",
+      "T(5,2) =  [7, 5]\n",
+      "If, T(x1,x2) = 0, then\n",
+      "x1 = x2 = 0\n",
+      "So, T is non-singular\n",
+      "z1,z2 are two scalars in F\n",
+      "z1 =  0\n",
+      "z2 =  8\n",
+      "So, x1 =  8\n",
+      "x2 =  -8\n",
+      "Hence, T is onto.\n",
+      "inverse(T) =  [8, -8]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "#x = round(rand(1,2) * 10)#\n",
+    "x1 = np.random.randint(1,9)\n",
+    "x2 = np.random.randint(1,9)\n",
+    "T = [x1+x2 ,x1]\n",
+    "print 'x1 = ',x1\n",
+    "print 'x2 = ',x2\n",
+    "print 'T(%d,%d) = '%(x1,x2),\n",
+    "print T\n",
+    "print 'If, T(x1,x2) = 0, then'\n",
+    "print 'x1 = x2 = 0'\n",
+    "print 'So, T is non-singular'\n",
+    "print 'z1,z2 are two scalars in F'\n",
+    "\n",
+    "z1 = np.random.randint(0,9)\n",
+    "z2 = np.random.randint(0,9)\n",
+    "print 'z1 = ',z1\n",
+    "print 'z2 = ',z2\n",
+    "x1 = z2#\n",
+    "x2 = z1 - z2#\n",
+    "print 'So, x1 = ',x1\n",
+    "print 'x2 = ',x2\n",
+    "print 'Hence, T is onto.'\n",
+    "Tinv = [z2, z1-z2]# \n",
+    "print 'inverse(T) = ',Tinv"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 89 Example 3.14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "T is a linear operator on F**2 defined as:\n",
+      "T(x1,x2) = (x1,0)\n",
+      "B = {e1,e2} is a standard ordered basis for F**2,then\n",
+      "So, Te1 = T(1,0) =  [1, 0]\n",
+      "So, Te2 = T(0,1) =  [0, 0]\n",
+      "so,matrix T in ordered basis B is: \n",
+      "T = \n",
+      "[[1, 0], [0, 0]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "print 'T is a linear operator on F**2 defined as:'\n",
+    "print 'T(x1,x2) = (x1,0)'\n",
+    "print 'B = {e1,e2} is a standard ordered basis for F**2,then'\n",
+    "x1 = 1#\n",
+    "x2 = 0#\n",
+    "Te1 = [x1, 0]#\n",
+    "x1 = 0#\n",
+    "x2 = 1#\n",
+    "Te2 = [x1 ,0]#\n",
+    "print 'So, Te1 = T(1,0) = ',Te1\n",
+    "print 'So, Te2 = T(0,1) = ',Te2\n",
+    "print 'so,matrix T in ordered basis B is: '\n",
+    "T = [Te1,Te2]\n",
+    "print 'T = \\n',T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 89 Example 3.15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Differentiation operator D is defined as:\n",
+      "(Df1)(x) =  0\n",
+      "(Df2)(x) =  1\n",
+      "(Df3)(x) =  2*x\n",
+      "(Df4)(x) =  3*x**2\n",
+      "Matrix of D in ordered basis is:\n",
+      "[D] = \n",
+      "[[ 0.  1.  0.  0.]\n",
+      " [ 0.  0.  2.  0.]\n",
+      " [ 0.  0.  0.  3.]\n",
+      " [ 0.  0.  0.  0.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "print 'Differentiation operator D is defined as:'\n",
+    "D = np.zeros([4,4])\n",
+    "x=sp.Symbol('x')\n",
+    "for i in range(1,5):\n",
+    "    t= i-1#\n",
+    "    f = sp.diff(x**t,'x')\n",
+    "    print '(Df%d)(x) = '%(i),\n",
+    "    print f\n",
+    "    if  not(i == 1):\n",
+    "        D[i-2,i-1] =  i-1#\n",
+    "    \n",
+    "\n",
+    "print 'Matrix of D in ordered basis is:'\n",
+    "print '[D] = \\n',D"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 92 Example 3.16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "T is a linear operator on R**2 defined as T(x1,x2) = (x1,0)\n",
+      "So, the matrix T in standard ordered basis B = {e1,e2} is \n",
+      "[T]B =  [[1 0]\n",
+      " [0 0]]\n",
+      "Let B is the ordered basis for R**2 consisting of vectors:\n",
+      "E1 =  [1 1]\n",
+      "E2 =  [2 1]\n",
+      "So, matrix P = \n",
+      "[[1 2]\n",
+      " [1 1]]\n",
+      "P inverse = \n",
+      "[[-1.  2.]\n",
+      " [ 1. -1.]]\n",
+      "So, matrix T in ordered basis B is [T]B = \n",
+      "[[-1.  0.]\n",
+      " [ 0. -0.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "print 'T is a linear operator on R**2 defined as T(x1,x2) = (x1,0)'\n",
+    "print 'So, the matrix T in standard ordered basis B = {e1,e2} is '\n",
+    "T = np.array([[1, 0],[0, 0]])\n",
+    "print '[T]B = ',T\n",
+    "print 'Let B'' is the ordered basis for R**2 consisting of vectors:'\n",
+    "E1 = np.array([1, 1])\n",
+    "E2 = np.array([2 ,1])\n",
+    "print 'E1 = ',E1\n",
+    "print 'E2 = ',E2\n",
+    "P = np.transpose(([E1,E2]))\n",
+    "print 'So, matrix P = \\n',P\n",
+    "Pinv=np.linalg.inv(P)\n",
+    "print 'P inverse = \\n',Pinv\n",
+    "T1 = Pinv*T*P#\n",
+    "print 'So, matrix T in ordered basis B'' is [T]B'' = \\n',T1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 93 Example 3.17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "g1 = f1\n",
+      "g2 = t*f1 + f2\n",
+      "g3 = t**2*f1 + 2*t*f2 + f3\n",
+      "g4 = t**3*f1 + 3*t**2*f2 + 3*t*f3 + f4\n",
+      "P = \n",
+      "Matrix([[1, t, t**2, t**3], [0, 1, 2*t, 3*t**2], [0, 0, 1, 3*t], [0, 0, 0, 1]])\n",
+      "inverse P = \n",
+      "Matrix([[1, -t, t**2, -t**3], [0, 1, -2*t, 3*t**2], [0, 0, 1, -3*t], [0, 0, 0, 1]])\n",
+      "Matrix of differentiation operator D in ordered basis B is:\n",
+      "D = \n",
+      "Matrix([[0, 1, 0, 0], [0, 0, 2, 0], [0, 0, 0, 3], [0, 0, 0, 0]])\n",
+      "Matrix of D in ordered basis B is:\n",
+      "inverse(P) * D * P =  Matrix([[0, 1, 0, 0], [0, 0, 2, 0], [0, 0, 0, 3], [0, 0, 0, 0]])\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "t = sp.Symbol(\"t\")\n",
+    "print 'g1 = f1'\n",
+    "print 'g2 = t*f1 + f2'\n",
+    "print 'g3 = t**2*f1 + 2*t*f2 + f3'\n",
+    "print 'g4 = t**3*f1 + 3*t**2*f2 + 3*t*f3 + f4'\n",
+    "P = sp.Matrix(([1, t, t**2, t**3],[0 ,1 ,2*t, 3*t**2],[0, 0, 1, 3*t],[0, 0, 0, 1]))\n",
+    "print 'P = \\n',P\n",
+    "\n",
+    "print 'inverse P = \\n',sp.Matrix.inv(P)\n",
+    "\n",
+    "\n",
+    "\n",
+    "print 'Matrix of differentiation operator D in ordered basis B is:'# #As found in example 15\n",
+    "D = sp.Matrix(([0, 1, 0, 0],[0, 0, 2, 0],[0, 0, 0, 3],[0, 0, 0, 0]))\n",
+    "print 'D = \\n',D\n",
+    "print 'Matrix of D in ordered basis B'' is:'\n",
+    "print 'inverse(P) * D * P = ',sp.Matrix.inv(P)*D*P\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 98 Example 3.19"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "n =  8\n",
+      "A = \n",
+      "[[  7.   3.   0.   4.   6.   8.   4.   4.]\n",
+      " [  6.   4.   2.   8.   7.   8.   1.   7.]\n",
+      " [  7.   0.   9.   3.  10.   9.   3.   8.]\n",
+      " [  7.   5.  10.   1.   8.   6.   6.   5.]\n",
+      " [  8.   8.   9.   9.   1.   9.  10.   4.]\n",
+      " [  6.   3.   5.   2.   2.   4.   8.   4.]\n",
+      " [  5.   1.   1.   2.   6.   9.   9.   5.]\n",
+      " [  8.   6.   9.   9.   8.   9.   1.   2.]]\n",
+      "Trace of A:\n",
+      "tr(A) =  37.0\n",
+      "--------------------------------\n",
+      "c =  3\n",
+      "B = \n",
+      "[[  4.   6.  10.   5.   8.   4.   1.   9.]\n",
+      " [  9.   9.   3.   6.   3.   8.   2.   6.]\n",
+      " [  1.   6.   0.   7.   7.   2.   8.   4.]\n",
+      " [  5.   5.   9.   7.   9.   3.   9.   9.]\n",
+      " [  7.   7.  10.   6.   1.   1.   7.   4.]\n",
+      " [  0.   3.  10.   9.   5.   2.   8.   4.]\n",
+      " [  1.   8.   2.   4.   5.   4.   4.   8.]\n",
+      " [  7.   0.   1.   8.   2.   7.   4.   7.]]\n",
+      "Trace of B:\n",
+      "tr(B) =  34.0\n",
+      "tr(cA + B) =  145.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "def trace_matrix(M,n):\n",
+    "    tr=0\n",
+    "    for i in range(0,n):\n",
+    "        tr = tr + M[i,i]#\n",
+    "    return tr\n",
+    "#n = round(rand() * 10 + 2)#\n",
+    "n=np.random.randint(1,9)\n",
+    "print 'n = ',n\n",
+    "#A = round(rand(n,n) * 10)#\n",
+    "A=np.random.rand(n,n)\n",
+    "for x in range(0,n):\n",
+    "    for y in range(0,n):\n",
+    "        A[x,y]=round(A[x,y]*10)\n",
+    "print 'A = \\n',A\n",
+    "\n",
+    "\n",
+    "tr = 0#\n",
+    "print 'Trace of A:'\n",
+    "tr1 = trace_matrix(A,n)#\n",
+    "print 'tr(A) = ',tr1\n",
+    "print '--------------------------------'\n",
+    "#c = round(rand() * 10 + 2)#\n",
+    "c=np.random.randint(2,9)\n",
+    "print 'c = ',c\n",
+    "\n",
+    "B=np.random.rand(n,n)\n",
+    "for x in range(0,n):\n",
+    "    for y in range(0,n):\n",
+    "        B[x,y]=round(B[x,y]*10)\n",
+    "print 'B = \\n',B\n",
+    "\n",
+    "print 'Trace of B:'\n",
+    "tr2 = trace_matrix(B,n)#\n",
+    "print 'tr(B) = ',tr2\n",
+    "print 'tr(cA + B) = ',(c*tr1+tr2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 103 Example 3.23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Matrix represented by given linear functionals on R**4:\n",
+      "A = \n",
+      "[[ 1  2  2  1]\n",
+      " [ 0  2  0  1]\n",
+      " [-2  0 -4  3]]\n",
+      "To find Row reduced echelon matrix of A given by R:\n",
+      "Applying row transformations on A,we get\n",
+      "R1 = R1-R2\n",
+      "A = \n",
+      "[[ 1  0  2  0]\n",
+      " [ 0  2  0  1]\n",
+      " [-2  0 -4  3]]\n",
+      "R3 = R3 + 2*R1\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 2 0 1]\n",
+      " [0 0 0 3]]\n",
+      "R3 = R3/3\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 2 0 1]\n",
+      " [0 0 0 1]]\n",
+      "R2 = R2/2\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 1 0 0]\n",
+      " [0 0 0 1]]\n",
+      "R2 = R2 - 1/2*R3\n",
+      "A = \n",
+      "[[1 0 2 0]\n",
+      " [0 1 0 0]\n",
+      " [0 0 0 1]]\n",
+      "Row reduced echelon matrix of A is:\n",
+      "R = \n",
+      "[[1 0 2 0]\n",
+      " [0 1 0 0]\n",
+      " [0 0 0 1]]\n",
+      "Therefore,linear functionals g1,g2,g3 span the same subspace of (R**4)* as f1,f2,f3 are given by:\n",
+      "g1(x1,x2,x3,x4) = x1 + 2*x3\n",
+      "g1(x1,x2,x3,x4) = x2\n",
+      "g1(x1,x2,x3,x4) = x4\n",
+      "The subspace consists of the vectors with\n",
+      "x1 = -2*x3\n",
+      "x2 = x4 = 0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "print 'Matrix represented by given linear functionals on R**4:'\n",
+    "A = np.array([[1, 2 ,2 ,1],[0, 2, 0, 1],[-2 ,0 ,-4, 3]])\n",
+    "print 'A = \\n',A\n",
+    "T = A              #Temporary matrix to store A\n",
+    "print 'To find Row reduced echelon matrix of A given by R:'\n",
+    "print 'Applying row transformations on A,we get'\n",
+    "print 'R1 = R1-R2'\n",
+    "A[0,:] = A[0,:] - A[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'R3 = R3 + 2*R1'\n",
+    "A[2,:] = A[2,:] + 2*A[0,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'R3 = R3/3'\n",
+    "A[2,:] = 1./3*A[2,:]#\n",
+    "print 'A = \\n',A\n",
+    "print 'R2 = R2/2'\n",
+    "A[1,:] = 1./2*A[1,:]\n",
+    "print 'A = \\n',A\n",
+    "print 'R2 = R2 - 1/2*R3'\n",
+    "A[1,:] = A[1,:] - 1./2*A[2,:]\n",
+    "print 'A = \\n',A\n",
+    "R = A#\n",
+    "A = T#\n",
+    "print 'Row reduced echelon matrix of A is:'\n",
+    "print 'R = \\n',R\n",
+    "print 'Therefore,linear functionals g1,g2,g3 span the same subspace of (R**4)* as f1,f2,f3 are given by:'\n",
+    "print 'g1(x1,x2,x3,x4) = x1 + 2*x3'\n",
+    "print 'g1(x1,x2,x3,x4) = x2'\n",
+    "print 'g1(x1,x2,x3,x4) = x4'\n",
+    "print 'The subspace consists of the vectors with'\n",
+    "print 'x1 = -2*x3'\n",
+    "print 'x2 = x4 = 0'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 104 Example 3.24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "W be the subspace of R**5 spanned by vectors:\n",
+      "a1 =  [2, -2, 3, 4, -1]\n",
+      "a2 =  [-1, 1, 2, 5, 2]\n",
+      "a3 =  [0, 0, -1, -2, 3]\n",
+      "a4 =  [1, -1, 2, 3, 0]\n",
+      "Matrix A by the row vectors a1,a2,a3,a4 will be:\n",
+      "A = \n",
+      "[[ 2 -2  3  4 -1]\n",
+      " [-1  1  2  5  2]\n",
+      " [ 0  0 -1 -2  3]\n",
+      " [ 1 -1  2  3  0]]\n",
+      "After Applying row transformations, we get the row reduced echelon matrix R of A\n",
+      "R = \n",
+      "[[ 1 -1  0 -1  0]\n",
+      " [ 0  0  1  2  0]\n",
+      " [ 0  0  0  0  1]\n",
+      " [ 0  0  0  0  0]]\n",
+      "Then we obtain all the linear functionals f by assigning arbitrary values to c2 and c4\n",
+      "Let c2 = a, c4 = b then c1 = a+b, c3 = -2b, c5 = 0.\n",
+      "So, W0 consists all linear functionals f of the form\n",
+      "f(x1,x2,x3,x4,x5) = (a+b)x1 + ax2 -2bx3 + bx4\n",
+      "Dimension of W0 = 2 and basis {f1,f2} can be found by first taking a = 1, b = 0. Then a = 0,b = 1\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "print 'W be the subspace of R**5 spanned by vectors:'\n",
+    "a1 = [2, -2, 3 ,4 ,-1]#\n",
+    "a2 = [-1, 1, 2, 5, 2]#\n",
+    "a3 = [0 ,0 ,-1, -2, 3]#\n",
+    "a4 = [1 ,-1, 2, 3 ,0]#\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print 'a4 = ',a4\n",
+    "print 'Matrix A by the row vectors a1,a2,a3,a4 will be:'\n",
+    "A = np.array([a1,a2,a3,a4])\n",
+    "print 'A = \\n',A\n",
+    "print 'After Applying row transformations, we get the row reduced echelon matrix R of A'\n",
+    "\n",
+    "T = A#                   #Temporary matrix to store A\n",
+    "#R1 = R1 - R4 and R2 = R2 + R4\n",
+    "A[0,:] = A[0,:] - A[3,:]#\n",
+    "A[1,:] = A[1,:] + A[3,:]#\n",
+    "#R2 = R2/2\n",
+    "A[1,:] = 1./2 * A[1,:]#\n",
+    "#R3 = R3 + R2 and R4 = R4 - R1\n",
+    "A[2,:] = A[2,:] + A[1,:]#\n",
+    "A[3,:] = A[3,:] - A[0,:]#\n",
+    "#R3 = R3 - R4\n",
+    "A[2,:] = A[2,:] - A[3,:]#\n",
+    "#R3 = R3/3\n",
+    "A[2,:] = 1./3 * A[2,:]#\n",
+    "#R2 = R2 - R3\n",
+    "A[1,:] = A[1,:] - A[2,:]#\n",
+    "#R2 = R2/2 and R4 = R4 - R2 - R3\n",
+    "A[1,:] = 1./2 * A[1,:]#\n",
+    "A[3,:] = A[3,:] - A[1,:] - A[2,:]#\n",
+    "#R1 = R1 - R2 + R3\n",
+    "A[0,:] = A[0,:] - A[1,:] + A[2,:]#\n",
+    "R = A#\n",
+    "A = T#\n",
+    "print 'R = \\n',R\n",
+    "print 'Then we obtain all the linear functionals f by assigning arbitrary values to c2 and c4'\n",
+    "print 'Let c2 = a, c4 = b then c1 = a+b, c3 = -2b, c5 = 0.'\n",
+    "print 'So, W0 consists all linear functionals f of the form'\n",
+    "print 'f(x1,x2,x3,x4,x5) = (a+b)x1 + ax2 -2bx3 + bx4'\n",
+    "print 'Dimension of W0 = 2 and basis {f1,f2} can be found by first taking a = 1, b = 0. Then a = 0,b = 1'"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter4_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter4_1.ipynb
new file mode 100644
index 00000000..3fb3b384
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter4_1.ipynb
@@ -0,0 +1,328 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 - Polynomials"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 121 Example 4.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "C is the field of complex numbers\n",
+      "f =  x**2 + 2\n",
+      "if a = C and z belongs to C, then f(z) = z**2 + 2\n",
+      "f(2) =  6\n",
+      "f(1+1J/1-1J) =  (1+0j)\n",
+      "----------------------------------------\n",
+      "If a is the algebra of all 2*2 matrices over C and\n",
+      "B = \n",
+      "[[ 1  0]\n",
+      " [-1  2]]\n",
+      "[[ 3.  0.]\n",
+      " [ 1.  6.]] then, f(B) = \n",
+      "----------------------------------------\n",
+      "If a is algebra of all linear operators on C**3\n",
+      "And T is element of a as:\n",
+      "T(c1,c2,c3) = (i*2**1/2*c1,c2,i*2**1/2*c3)\n",
+      "Then, f(T)(c1,c2,c3) = (0,3*c2,0)\n",
+      "----------------------------------------\n",
+      "If a is the algebra of all polynomials over C\n",
+      "And, g =  x**4 + 3.0*I\n",
+      "Then f(g) =  (16+3j)\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "from sympy.polys.polyfuncs import horner\n",
+    "print 'C is the field of complex numbers'\n",
+    "x = sp.Symbol(\"x\")\n",
+    "def f(x):\n",
+    "    ff= x**2 + 2\n",
+    "    return ff\n",
+    "print 'f = ',f(x)\n",
+    "#part a\n",
+    "print 'if a = C and z belongs to C, then f(z) = z**2 + 2'\n",
+    "print 'f(2) = ',f(2)\n",
+    "\n",
+    "\n",
+    "print 'f(1+1J/1-1J) = ',f((1+1J)/(1-1J))\n",
+    "print '----------------------------------------'\n",
+    "\n",
+    "\n",
+    "#part b\n",
+    "print 'If a is the algebra of all 2*2 matrices over C and'\n",
+    "B = np.array([[1 ,0],[-1, 2]])\n",
+    "print 'B = \\n',B\n",
+    "print 2*np.identity(2) + B**2,'then, f(B) = '\n",
+    "print '----------------------------------------'\n",
+    "\n",
+    "#part c\n",
+    "print 'If a is algebra of all linear operators on C**3'\n",
+    "print 'And T is element of a as:'\n",
+    "print 'T(c1,c2,c3) = (i*2**1/2*c1,c2,i*2**1/2*c3)'\n",
+    "print 'Then, f(T)(c1,c2,c3) = (0,3*c2,0)'\n",
+    "print '----------------------------------------'\n",
+    "#part d\n",
+    "print 'If a is the algebra of all polynomials over C'\n",
+    "def g(x):\n",
+    "    gg= x**4 + 3*1J\n",
+    "    return gg\n",
+    "print 'And, g = ',g(x)\n",
+    "print 'Then f(g) = ',g(2)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 131 Example 4.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "M = (x+2)F[x] + (x**2 + 8x + 16)F[x]\n",
+      "We assert, M = F[x]\n",
+      "M contains: x**2 - x*(x + 2) + 8*x + 16\n",
+      "And hence M contains: x**2 - x*(x + 2) + 2*x + 4\n",
+      "Thus the scalar polynomial 1 belongs to M as well all its multiples.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "x = sp.Symbol('x')\n",
+    "p1 = x + 2#\n",
+    "p2 = x**2 + 8*x + 16#\n",
+    "print 'M = (x+2)F[x] + (x**2 + 8x + 16)F[x]'\n",
+    "print 'We assert, M = F[x]'\n",
+    "print 'M contains:',\n",
+    "t = p2 - x*p1#\n",
+    "print t\n",
+    "print 'And hence M contains:',\n",
+    "print t - 6*p1\n",
+    "print 'Thus the scalar polynomial 1 belongs to M as well all its multiples.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 133 Example 4.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "p1 =  x + 2\n",
+      "p2 =  x**2 + 8*x + 16\n",
+      "M = (x+2)F[x] + (x**2 + 8x + 16)F[x]\n",
+      "We assert, M = F[x]\n",
+      "M contains: x**2 - x*(x + 2) + 8*x + 16\n",
+      "And hence M contains: x**2 - x*(x + 2) + 2*x + 4\n",
+      "Thus the scalar polynomial 1 belongs to M as well all its multiples\n",
+      "So, gcd(p1,p2) = 1\n",
+      "----------------------------------------------\n",
+      "p1 =  (x - 2)**2*(x + 1.0*I)\n",
+      "p2 =  (x - 2)*(x**2 + 1)\n",
+      "M = (x - 2)**2*(x+1J)F[x] + (x-2)*(x**2 + 1\n",
+      "The ideal M contains p1 - p2 i.e., (x - 2)**2*(x + 1.0*I) - (x - 2)*(x**2 + 1)\n",
+      "Hence it contains (x-2)(x+i), which is monic and divides both,\n",
+      "So, gcd(p1,p2) = (x-2)(x+i)\n",
+      "----------------------------------------------\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "x = sp.Symbol('x')\n",
+    "\n",
+    "#part a\n",
+    "p1 = x + 2#\n",
+    "p2 = x**2 + 8*x + 16#\n",
+    "print 'p1 = ',p1\n",
+    "print 'p2 = ',p2\n",
+    "print 'M = (x+2)F[x] + (x**2 + 8x + 16)F[x]'\n",
+    "print 'We assert, M = F[x]'\n",
+    "print 'M contains:',\n",
+    "t = p2 - x*p1#\n",
+    "print t\n",
+    "print 'And hence M contains:',\n",
+    "print t - 6*p1\n",
+    "print 'Thus the scalar polynomial 1 belongs to M as well all its multiples'\n",
+    "print 'So, gcd(p1,p2) = 1'\n",
+    "print '----------------------------------------------'\n",
+    "#part b\n",
+    "p1 = (x - 2)**2*(x+1J)#\n",
+    "p2 = (x-2)*(x**2 + 1)#\n",
+    "print 'p1 = ',p1\n",
+    "print 'p2 = ',p2\n",
+    "print 'M = (x - 2)**2*(x+1J)F[x] + (x-2)*(x**2 + 1'\n",
+    "print 'The ideal M contains p1 - p2 i.e.,',\n",
+    "print p1 - p2\n",
+    "print 'Hence it contains (x-2)(x+i), which is monic and divides both,'\n",
+    "print 'So, gcd(p1,p2) = (x-2)(x+i)'\n",
+    "print '----------------------------------------------'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 133 Example 4.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "M is the ideal in F[x] generated by:\n",
+      "(x-1)*(x+2)**2\n",
+      "(x+2)**2*(x+3)\n",
+      "(x-3) and\n",
+      "M = (x-1)*(x+2)**2 F[x] + (x+2)**2*(x-3) + (x-3)\n",
+      "Then M contains: 0\n",
+      "i.e., M contains (x+2)**2\n",
+      "and since, (x+2)**2 = (x-3)(x-7) - 17\n",
+      "So M contains the scalar polynomial 1.\n",
+      "So, M = F[x] and given polynomials are relatively prime.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "x = sp.Symbol('x')\n",
+    "\n",
+    "print 'M is the ideal in F[x] generated by:'\n",
+    "print '(x-1)*(x+2)**2'\n",
+    "print '(x+2)**2*(x+3)'\n",
+    "print '(x-3)','and'\n",
+    "p1 = (x-1)*(x+2)**2#\n",
+    "p2 = (x+2)**2*(x-3)#\n",
+    "p3 = (x-3)#\n",
+    "print 'M = (x-1)*(x+2)**2 F[x] + (x+2)**2*(x-3) + (x-3)'\n",
+    "print 'Then M contains:',\n",
+    "t = 1/2*(x+2)**2*((x-1) - (x-3))#\n",
+    "print t\n",
+    "print 'i.e., M contains (x+2)**2'\n",
+    "print 'and since, (x+2)**2 = (x-3)(x-7) - 17'\n",
+    "print 'So M contains the scalar polynomial 1.'\n",
+    "print 'So, M = F[x] and given polynomials are relatively prime.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 135 Example 4.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x**2 + 1 P = \n",
+      "P is reducible over complex numbers as:  = x**2 + 1\n",
+      "(x-i)(x+i)\n",
+      "Whereas, P is irreducible over real numbers as: = x**2 + 1\n",
+      "(ax + b)(ax + b)\n",
+      "For, a,a,b,b to be in R,\n",
+      "aa = 1\n",
+      "ab + ba = 0\n",
+      "bb = 1\n",
+      "=> a**2 + b**2 = 0\n",
+      "=> a = b = 0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "x = sp.Symbol('x')\n",
+    "P = x**2 + 1#\n",
+    "print P,'P = '\n",
+    "print 'P is reducible over complex numbers as: ',\n",
+    "print '=',P\n",
+    "print '(x-i)(x+i)'\n",
+    "print 'Whereas, P is irreducible over real numbers as:',\n",
+    "print '=',P\n",
+    "print '(ax + b)(a''x + b'')'\n",
+    "print 'For, a,a'',b,b'' to be in R,'\n",
+    "print 'aa'' = 1'\n",
+    "print 'ab'' + ba'' = 0'\n",
+    "print 'bb'' = 1'\n",
+    "print '=> a**2 + b**2 = 0'\n",
+    "print '=> a = b = 0'"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter5_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter5_1.ipynb
new file mode 100644
index 00000000..9d29340c
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter5_1.ipynb
@@ -0,0 +1,358 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 - Determinants"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 143 Example 5.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[ 3.  9.]\n",
+      " [ 3.  1.]]\n",
+      "D1(A) =  3.0\n",
+      "D2(A) =  -27.0\n",
+      "D(A) = D1(A) + D2(A) =  -24.0\n",
+      "That is, D is a 2-linear function.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "A = np.random.rand(2,2)\n",
+    "for x in range(0,2):\n",
+    "    for y in range(0,2):\n",
+    "        A[x,y]=round(A[x,y]*10)\n",
+    "print 'A = \\n',A\n",
+    "\n",
+    "D1 = A[0,0]*A[1,1]\n",
+    "D2 = - A[0,1]*A[1,0]\n",
+    "print 'D1(A) = ',D1\n",
+    "print 'D2(A) = ',D2\n",
+    "print 'D(A) = D1(A) + D2(A) = ',D1 + D2\n",
+    "print 'That is, D is a 2-linear function.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 145 Example 5.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "Matrix([[x, 0, -x**2], [0, 1, 0], [1, 0, x**3]])\n",
+      "e1,e2,e3 are the rows of 3*3 identity matrix, then\n",
+      "e1 =  [ 1.  0.  0.]\n",
+      "e2 =  [ 0.  1.  0.]\n",
+      "e3 =  [ 0.  0.  1.]\n",
+      "D(A) = D(x*e1 - x**2*e3, e2, e1 + x**3*e3)\n",
+      "Since, D is linear as a function of each row,\n",
+      "D(A) = x*D(e1,e2,e1 + x**3*e3) - x**2*D(e3,e2,e1 + x**3*e3)\n",
+      "D(A) = x*D(e1,e2,e1) + x**4*D(e1,e2,e3) - x**2*D(e3,e2,e1) - x**5*D(e3,e2,e3)\n",
+      "As D is alternating, So\n",
+      "D(A) = (x**4 + x**2)*D(e1,e2,e3)\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "\n",
+    "x = sp.Symbol(\"x\")\n",
+    "A = sp.Matrix(([x, 0, -x**2],[0, 1, 0],[1, 0, x**3]))\n",
+    "print 'A = \\n',A\n",
+    "print 'e1,e2,e3 are the rows of 3*3 identity matrix, then'\n",
+    "T = np.identity(3)\n",
+    "e1 = T[0,:]\n",
+    "e2 = T[1,:]\n",
+    "e3 = T[2,:]\n",
+    "print 'e1 = ',e1\n",
+    "print 'e2 = ',e2\n",
+    "print 'e3 = ',e3\n",
+    "print 'D(A) = D(x*e1 - x**2*e3, e2, e1 + x**3*e3)'\n",
+    "print 'Since, D is linear as a function of each row,'\n",
+    "print 'D(A) = x*D(e1,e2,e1 + x**3*e3) - x**2*D(e3,e2,e1 + x**3*e3)'\n",
+    "print 'D(A) = x*D(e1,e2,e1) + x**4*D(e1,e2,e3) - x**2*D(e3,e2,e1) - x**5*D(e3,e2,e3)'\n",
+    "print 'As D is alternating, So'\n",
+    "print 'D(A) = (x**4 + x**2)*D(e1,e2,e3)'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 147 Example 5.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "Matrix([[x - 1, x**2, x**3], [0, x - 2, 1], [0, 0, x - 3]])\n",
+      "\n",
+      "E(A) =  x**3 - 6*x**2 + 11*x - 6\n",
+      "--------------------------------------\n",
+      "A = \n",
+      "Matrix([[0, 1, 0], [0, 0, 1], [1, 0, 0]])\n",
+      "\n",
+      "E(A) =  1\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "\n",
+    "#part a\n",
+    "x = sp.Symbol('x')\n",
+    "A = sp.Matrix(([x-1, x**2, x**3],[0, x-2, 1],[0, 0, x-3]))\n",
+    "print 'A = \\n',A\n",
+    "print '\\nE(A) = ',A.det()\n",
+    "print '--------------------------------------'\n",
+    "#part b\n",
+    "A = sp.Matrix(([0 ,1, 0],[0, 0, 1],[1 ,0, 0]))\n",
+    "print 'A = \\n',A\n",
+    "print '\\nE(A) = ',A.det()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 158 Example 5.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Given Matrix:\n",
+      "A = \n",
+      "[[ 1 -1  2  3]\n",
+      " [ 2  2  0  2]\n",
+      " [ 4  1 -1 -1]\n",
+      " [ 1  2  3  0]]\n",
+      "After, Subtracting muliples of row 1 from rows 2 3 4\n",
+      "R2 = R2 - 2*R1\n",
+      "R3 = R3 - 4*R1\n",
+      "R4 = R4 - R1\n",
+      "A = \n",
+      "[[  1  -1   2   3]\n",
+      " [  0   4  -4  -4]\n",
+      " [  0   5  -9 -13]\n",
+      " [  0   3   1  -3]]\n",
+      "We obtain the same determinant as before.\n",
+      "Now, applying some more row transformations as:\n",
+      "R3 = R3 - 5/4 * R2\n",
+      "R4 = R4 - 3/4 * R2\n",
+      "We get B as:\n",
+      "B = \n",
+      "[[ 1 -1  2  3]\n",
+      " [ 0  4 -4 -4]\n",
+      " [ 0  0 -4 -8]\n",
+      " [ 0  0  4  0]]\n",
+      "Now,determinant of A and B will be same\n",
+      "det A = det B =  128.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "print 'Given Matrix:'\n",
+    "A = np.array([[1, -1, 2, 3],[ 2, 2, 0, 2],[ 4, 1 ,-1, -1],[1, 2, 3, 0]])\n",
+    "print 'A = \\n',A\n",
+    "print 'After, Subtracting muliples of row 1 from rows 2 3 4'\n",
+    "print 'R2 = R2 - 2*R1'\n",
+    "A[1,:] = A[1,:] - 2 * A[0,:]\n",
+    "print 'R3 = R3 - 4*R1'\n",
+    "A[2,:] = A[2,:] - 4 * A[0,:]\n",
+    "print 'R4 = R4 - R1'\n",
+    "A[3,:] = A[3,:] - A[0,:]\n",
+    "print 'A = \\n',A\n",
+    "T = A#                  #Temporary matrix to store A\n",
+    "print 'We obtain the same determinant as before.'\n",
+    "print 'Now, applying some more row transformations as:'\n",
+    "print 'R3 = R3 - 5/4 * R2'\n",
+    "T[2,:] = T[2,:] - 5./4 * T[1,:]\n",
+    "print 'R4 = R4 - 3/4 * R2'\n",
+    "T[3,:] = T[3,:] - 3./4 * T[1,:]\n",
+    "B = T#\n",
+    "print 'We get B as:'\n",
+    "print 'B = \\n',B\n",
+    "print 'Now,determinant of A and B will be same'\n",
+    "print 'det A = det B = ',np.linalg.det(B)\n",
+    "     "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 160 Example 5.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "Matrix([[x**2 + x, x + 1], [x - 1, 1]])\n",
+      "B = \n",
+      "Matrix([[x**2 - 1, x + 2], [x**2 - 2*x + 3, x]])\n",
+      "det A =  x + 1\n",
+      "det B =  -6\n",
+      "Thus, A is not invertible over K whereas B is invertible\n",
+      "adj A =  Matrix([[(x + 1)*((x - 1)/(x*(x + 1)) + 1/(x*(x + 1))), -x - 1], [-x + 1, x*(x + 1)]])\n",
+      "adj B =  Matrix([[-6*(x + 2)*(-x**2/6 + 1/6)*(x**2 - 2*x + 3)/(x**2 - 1)**2 - 6/(x**2 - 1), 6*(x + 2)*(-x**2/6 + 1/6)/(x**2 - 1)], [6*(-x**2/6 + 1/6)*(x**2 - 2*x + 3)/(x**2 - 1), x**2 - 1]])\n",
+      "(adj A)A = (x+1)I\n",
+      "(adj B)B =  -6I\n",
+      "B inverse =  Matrix([[(x + 2)*(-x**2/6 + 1/6)*(x**2 - 2*x + 3)/(x**2 - 1)**2 + 1/(x**2 - 1), -(x + 2)*(-x**2/6 + 1/6)/(x**2 - 1)], [-(-x**2/6 + 1/6)*(x**2 - 2*x + 3)/(x**2 - 1), -x**2/6 + 1/6]])\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "import numpy as np\n",
+    "\n",
+    "x = sp.Symbol(\"x\")\n",
+    "A = sp.Matrix(([x**2+x, x+1],[x-1, 1]))\n",
+    "B = sp.Matrix(([x**2-1, x+2],[x**2-2*x+3, x]))\n",
+    "print 'A = \\n',A\n",
+    "print 'B = \\n',B\n",
+    "print 'det A = ',A.det()\n",
+    "print 'det B = ',B.det()\n",
+    "print 'Thus, A is not invertible over K whereas B is invertible'\n",
+    "\n",
+    "print 'adj A = ',(A**-1)*A.det()\n",
+    "print 'adj B = ',(B**-1)*B.det()\n",
+    "print '(adj A)A = (x+1)I'\n",
+    "print '(adj B)B =  -6I'\n",
+    "print 'B inverse = ',B**-1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 161 Example 5.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[1 2]\n",
+      " [3 4]]\n",
+      "det A =  Determinant of A is: -2.0\n",
+      "Adjoint of A is: [[-2.  -0. ]\n",
+      " [-0.  -0.5]]\n",
+      "Thus, A is not invertible as a matrix over the ring of integers.\n",
+      "But, A can be regarded as a matrix over field of rational numbers.\n",
+      "Then, A is invertible and Inverse of A is: [[-2.   1. ]\n",
+      " [ 1.5 -0.5]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "A = np.array([[1, 2],[3, 4]])\n",
+    "print 'A = \\n',A\n",
+    "d = np.linalg.det(A)#\n",
+    "print 'det A = ','Determinant of A is:',d\n",
+    "\n",
+    "\n",
+    "ad = (d* np.identity(2)) / A\n",
+    "print 'Adjoint of A is:',ad\n",
+    "\n",
+    "\n",
+    "print 'Thus, A is not invertible as a matrix over the ring of integers.'\n",
+    "print 'But, A can be regarded as a matrix over field of rational numbers.'\n",
+    "In = np.linalg.inv(A)#\n",
+    "#The A inverse matrix given in book has a wrong entry of 1/2. It should be -1/2.\n",
+    "print 'Then, A is invertible and Inverse of A is:',In\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter6_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter6_1.ipynb
new file mode 100644
index 00000000..9ad1cbe2
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter6_1.ipynb
@@ -0,0 +1,503 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 6 - Elementary canonical forms"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 184 Example 6.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Standard ordered matrix for Linear operator T on R**2 is:\n",
+      "A = \n",
+      "Matrix([[0, -1], [1, 0]])\n",
+      "The characteristic polynomial for T or A is: Matrix([[x, 1], [-1, x]])\n",
+      "Since this polynomial has no real roots,T has no characteristic values.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "print 'Standard ordered matrix for Linear operator T on R**2 is:'\n",
+    "A = sp.Matrix(([0, -1],[1 ,0]))\n",
+    "print 'A = \\n',A\n",
+    "print 'The characteristic polynomial for T or A is:',\n",
+    "x = sp.Symbol(\"x\")\n",
+    "p = (x*sp.eye(2)-A)\n",
+    "print p\n",
+    "print 'Since this polynomial has no real roots,T has no characteristic values.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 184 Example 6.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "Matrix([[3, 1, -1], [2, 2, -1], [2, 2, 0]])\n",
+      "Characteristic polynomial for A is: x**3 - 5*x**2 + 8*x - 4\n",
+      "or\n",
+      "(x-1)(x-2)**2\n",
+      "The characteristic values of A are:\n",
+      "[1, 2]\n",
+      "Now, A-I = \n",
+      "Matrix([[2, 1, -1], [2, 1, -1], [2, 2, -1]])\n",
+      "rank of A - I=  2\n",
+      "So, nullity of T-I = 1\n",
+      "The vector that spans the null space of T-I =  [1, 0, 2]\n",
+      "Now,A-2I = \n",
+      "Matrix([[1, 1, -1], [2, 0, -1], [2, 2, -2]])\n",
+      "rank of A - 2I=  2\n",
+      "T*alpha = 2*alpha if alpha is a scalar multiple of a2\n",
+      "a2 =  [1, 1, 2]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "A = sp.Matrix(([3, 1, -1],[ 2, 2, -1],[2, 2, 0]))\n",
+    "print 'A = \\n',A\n",
+    "print 'Characteristic polynomial for A is:',\n",
+    "x=sp.Symbol('x')\n",
+    "p = A.charpoly(x)#\n",
+    "print p.as_expr()\n",
+    "print 'or'\n",
+    "print '(x-1)(x-2)**2'\n",
+    "\n",
+    "r = sp.solve(p.as_expr())#\n",
+    "[m,n] = A.shape\n",
+    "print 'The characteristic values of A are:'\n",
+    "print r  #print round(r)\n",
+    "B = A-sp.eye(m)\n",
+    "print 'Now, A-I = \\n',B\n",
+    "\n",
+    "print 'rank of A - I= ',B.rank()\n",
+    "print 'So, nullity of T-I = 1'\n",
+    "a1 = [1 ,0 ,2]#\n",
+    "print 'The vector that spans the null space of T-I = ',a1\n",
+    "B = A-2*sp.eye(m)\n",
+    "print 'Now,A-2I = \\n',B\n",
+    "print 'rank of A - 2I= ',B.rank()\n",
+    "print 'T*alpha = 2*alpha if alpha is a scalar multiple of a2'\n",
+    "a2 = [1 ,1 ,2]\n",
+    "print 'a2 = ',a2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 187 Example 6.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Standard ordered matrix for Linear operator T on R**3 is:\n",
+      "A = \n",
+      "Matrix([[5, -6, -6], [-1, 4, 2], [3, -6, -4]])\n",
+      "xI - A = \n",
+      "Matrix([[x - 5, 6, 6], [1, x - 4, -2], [-3, 6, x + 4]])\n",
+      "Applying row and column transformations:\n",
+      "C2 = C2 - C3\n",
+      "=>\n",
+      "Matrix([[x - 5, 0, 6], [1, x - 2, -2], [-3, -x + 2, x + 4]])\n",
+      "Taking (x-2) common from C2\n",
+      "=>\n",
+      " *  x - 2\n",
+      "Matrix([[x - 5, 0, 6], [1, 1, -2], [-3, (-x + 2)/(x - 2), x + 4]])\n",
+      "R3 = R3 + R2\n",
+      "=>\n",
+      " *  x - 2\n",
+      "Matrix([[x - 5, 0, 6], [1, 1, -2], [-2, (-x + 2)/(x - 2) + 1, x + 2]])\n",
+      "=>\n",
+      " *  x - 2\n",
+      "Matrix([[x - 5, 6], [-2, x + 2]])\n",
+      "=>\n",
+      " *  x - 2\n",
+      "x**2 - 3*x + 2\n",
+      "This is the characteristic polynomial\n",
+      "Now, A - I =  Matrix([[4, -6, -6], [-1, 3, 2], [3, -6, -5]])\n",
+      "And, A- 2I =  Matrix([[3, -6, -6], [-1, 2, 2], [3, -6, -6]])\n",
+      "rank(A-I) =  2\n",
+      "rank(A-2I) =  2\n",
+      "W1,W2 be the spaces of characteristic vectors associated with values 1,2\n",
+      "So by theorem 2, T is diagonalizable\n",
+      "Null space of (T- I) i.e basis of W1 is spanned by a1 =  [[ 3 -1  3]]\n",
+      "Null space of (T- 2I) i.e. basis of W2 is spanned by vectors x1,x2,x3 such that x1 = 2x1 + 2x3\n",
+      "One example :\n",
+      "a2 =  [[2 1 0]]\n",
+      "a3 =  [[2 0 1]]\n",
+      "The diagonal matrix is:\n",
+      "D =  [[1 0 0]\n",
+      " [0 2 0]\n",
+      " [0 0 2]]\n",
+      "The standard basis matrix is denoted as:\n",
+      "P =  [[ 3  2  2]\n",
+      " [-1  1  0]\n",
+      " [ 3  0  1]]\n",
+      "AP =  Matrix([[3, 4, 4], [-1, 2, 0], [3, 0, 2]])\n",
+      "PD =  [[3 0 0]\n",
+      " [0 2 0]\n",
+      " [0 0 2]]\n",
+      "That is, AP = PD\n",
+      "=>  inverse(P)*A*P = D\n"
+     ]
+    }
+   ],
+   "source": [
+    "import sympy as sp\n",
+    "import numpy as np\n",
+    "print 'Standard ordered matrix for Linear operator T on R**3 is:'\n",
+    "A = sp.Matrix(([5, -6, -6],[ -1, 4, 2],[ 3, -6, -4]))\n",
+    "print 'A = \\n',A\n",
+    "print 'xI - A = '\n",
+    "B = sp.eye(3)\n",
+    "x = sp.Symbol('x')\n",
+    "P = x*B - A#\n",
+    "print P\n",
+    "\n",
+    "print 'Applying row and column transformations:'\n",
+    "print 'C2 = C2 - C3'\n",
+    "P[:,1] = P[:,1] - P[:,2]\n",
+    "print '=>'\n",
+    "print P\n",
+    "print 'Taking (x-2) common from C2'\n",
+    "c = x-2#\n",
+    "P[:,1] = P[:,1] / (x-2)\n",
+    "print '=>'\n",
+    "print ' * ', c\n",
+    "print P\n",
+    "print 'R3 = R3 + R2'\n",
+    "P[2,:] = P[2,:] + P[1,:]\n",
+    "print '=>'\n",
+    "print ' * ', c\n",
+    "print P\n",
+    "P = sp.Matrix(([P[0,0], P[0,2]],[P[2,0], P[2,2]]))\n",
+    "print '=>'\n",
+    "print ' * ', c\n",
+    "print P\n",
+    "print '=>'\n",
+    "print ' * ',c\n",
+    "print P.det()\n",
+    "print 'This is the characteristic polynomial'\n",
+    "\n",
+    "print 'Now, A - I = ',A-B\n",
+    "print 'And, A- 2I = ',A-2*B\n",
+    "print 'rank(A-I) = ',np.rank(A-B)\n",
+    "\n",
+    "print 'rank(A-2I) = ',np.rank(A-2*B)\n",
+    "print 'W1,W2 be the spaces of characteristic vectors associated with values 1,2'\n",
+    "print 'So by theorem 2, T is diagonalizable'\n",
+    "a1 = np.array([[3, -1 ,3]])\n",
+    "a2 = np.array([[2, 1, 0]])\n",
+    "a3 = np.array([[2, 0, 1]])\n",
+    "print 'Null space of (T- I) i.e basis of W1 is spanned by a1 = ',a1\n",
+    "print 'Null space of (T- 2I) i.e. basis of W2 is spanned by vectors x1,x2,x3 such that x1 = 2x1 + 2x3'\n",
+    "print 'One example :'\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print 'The diagonal matrix is:'\n",
+    "D = np.array([[1 ,0 ,0 ],[0, 2, 0],[0, 0, 2]])\n",
+    "print 'D = ',D\n",
+    "print 'The standard basis matrix is denoted as:'\n",
+    "P = np.transpose(np.vstack([a1,a2,a3]))\n",
+    "print 'P = ',P\n",
+    "print 'AP = ',A*P\n",
+    "print 'PD = ',P*D\n",
+    "print 'That is, AP = PD'\n",
+    "print '=>  inverse(P)*A*P = D'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 193 Example 6.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[ 5 -6 -6]\n",
+      " [-1  4  2]\n",
+      " [ 3 -6 -4]]\n",
+      "Characteristic polynomial of A is:\n",
+      "f = (x-1)(x-2)**2\n",
+      "i.e., f =  (x - 2)**2*(x - 1)\n",
+      "(A-I)(A-2I) =  Matrix([[0, 0, 0], [0, 0, 0], [0, 0, 0]])\n",
+      "Since, (A-I)(A-2I) = 0. So, Minimal polynomial for above is:\n",
+      "p =  (x - 2)*(x - 1)\n",
+      "---------------------------------------\n",
+      "A = \n",
+      "[[ 3  1 -1]\n",
+      " [ 2  2 -1]\n",
+      " [ 2  2  0]]\n",
+      "Characteristic polynomial of A is:\n",
+      "f = (x-1)(x-2)**2\n",
+      "i.e., f =  (x - 2)**2*(x - 1)\n",
+      "(A-I)(A-2I) =  Matrix([[2, 0, -1], [2, 0, -1], [4, 0, -2]])\n",
+      "Since, (A-I)(A-2I) is not equal to 0. T is not diagonalizable. So, Minimal polynomial cannot be p.\n",
+      "---------------------------------------\n",
+      "A = \n",
+      "[[ 0 -1]\n",
+      " [ 1  0]]\n",
+      "Characteristic polynomial of A is:\n",
+      "f =  x**2 + 1\n",
+      "A**2 + I =  Matrix([[1, 1], [1, 1]])\n",
+      "Since, A**2 + I = 0, so minimal polynomial is\n",
+      "p =  x**2 + 1\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "\n",
+    "x = sp.Symbol(\"x\")\n",
+    "A = np.array([[5, -6, -6],[ -1, 4 ,2],[ 3, -6, -4]])   #Matrix given in Example 3\n",
+    "print 'A = \\n',A\n",
+    "f = (x-1)*(x-2)**2# \n",
+    "print 'Characteristic polynomial of A is:'\n",
+    "print 'f = (x-1)(x-2)**2'\n",
+    "print 'i.e., f = ',f\n",
+    "p = (x-1)*(x-2)#\n",
+    "print '(A-I)(A-2I) = ',(A-sp.eye(3))*(A-2 * sp.eye(3))\n",
+    "print 'Since, (A-I)(A-2I) = 0. So, Minimal polynomial for above is:'\n",
+    "print 'p = ',p\n",
+    "print '---------------------------------------'\n",
+    "\n",
+    "A = np.array([[3, 1 ,-1],[ 2, 2 ,-1],[2, 2, 0]])    #Matrix given in Example 2\n",
+    "print 'A = \\n',A\n",
+    "f = (x-1)*(x-2)**2# \n",
+    "print 'Characteristic polynomial of A is:'\n",
+    "print 'f = (x-1)(x-2)**2'\n",
+    "print 'i.e., f = ',f\n",
+    "print '(A-I)(A-2I) = ',(A-sp.eye(3))*(A-2 * sp.eye(3))\n",
+    "print 'Since, (A-I)(A-2I) is not equal to 0. T is not diagonalizable. So, Minimal polynomial cannot be p.'\n",
+    "print '---------------------------------------'\n",
+    "A = np.array([[0, -1],[1, 0]])\n",
+    "print 'A = \\n',A\n",
+    "f = x**2 + 1#\n",
+    "print 'Characteristic polynomial of A is:'\n",
+    "print 'f = ',f\n",
+    "print 'A**2 + I = ',A**2 + sp.eye(2)\n",
+    "print 'Since, A**2 + I = 0, so minimal polynomial is'\n",
+    "p = x**2 + 1\n",
+    "print 'p = ',p"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 197 Example 6.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[0 1 0 1]\n",
+      " [1 0 1 0]\n",
+      " [0 1 0 1]\n",
+      " [1 0 1 0]]\n",
+      "Computing powers on A:\n",
+      "A**2 = \n",
+      "[[0 1 0 1]\n",
+      " [1 0 1 0]\n",
+      " [0 1 0 1]\n",
+      " [1 0 1 0]]\n",
+      "A**3 = \n",
+      "[[0 1 0 1]\n",
+      " [1 0 1 0]\n",
+      " [0 1 0 1]\n",
+      " [1 0 1 0]]\n",
+      "if p = x**3 - 4x, then\n",
+      "p(A) =  [[ 0 -3  0 -3]\n",
+      " [-3  0 -3  0]\n",
+      " [ 0 -3  0 -3]\n",
+      " [-3  0 -3  0]]\n",
+      "Minimal polynomial for A is:  x**3 - 4*x\n",
+      "Characteristic values for A are: [-2, 0, 2]\n",
+      "Rank(A) =  2\n",
+      "So, from theorem 2, characteristic polynomial for A is: x**4 - 4*x**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "A = np.array([[0, 1, 0, 1],[1, 0 ,1 ,0],[0, 1, 0, 1],[1, 0, 1, 0]])\n",
+    "print 'A = \\n',A\n",
+    "print 'Computing powers on A:'\n",
+    "print 'A**2 = \\n',A*A\n",
+    "print 'A**3 = \\n',A*A*A\n",
+    "def p(x):\n",
+    "    pp = x**3 - 4*x\n",
+    "    return pp\n",
+    "print 'if p = x**3 - 4x, then'\n",
+    "print 'p(A) = ',p(A)\n",
+    "x = sp.Symbol(\"x\")\n",
+    "f = x**3 - 4*x\n",
+    "print 'Minimal polynomial for A is: ',f\n",
+    "print 'Characteristic values for A are:',sp.solve(f,x)\n",
+    "print 'Rank(A) = ',np.rank(A)\n",
+    "print 'So, from theorem 2, characteristic polynomial for A is:',sp.Matrix(A).charpoly(x).as_expr()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 210 Example 6.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "[[  9.   1.   3.]\n",
+      " [ 10.   1.   3.]\n",
+      " [ 10.   5.   1.]]\n",
+      "A transpose is:\n",
+      "A' = \n",
+      "[[  9.  10.  10.]\n",
+      " [  1.   1.   5.]\n",
+      " [  3.   3.   1.]]\n",
+      "Since, A' is not equal to A, A is not a symmetric matrix.\n",
+      "Since, A' is not equal to -A, A is not a skew-symmetric matrix.\n",
+      "A can be expressed as sum of A1 and A2\n",
+      "i.e., A = A1 + A2\n",
+      "A1 = \n",
+      "[[ 9.   5.5  6.5]\n",
+      " [ 5.5  1.   4. ]\n",
+      " [ 6.5  4.   1. ]]\n",
+      "A2 = \n",
+      "[[ 0.  -4.5 -3.5]\n",
+      " [ 4.5  0.  -1. ]\n",
+      " [ 3.5  1.   0. ]]\n",
+      "A1 + A2 = \n",
+      "[[  9.   1.   3.]\n",
+      " [ 10.   1.   3.]\n",
+      " [ 10.   5.   1.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "\n",
+    "A = np.random.rand(3,3)\n",
+    "for i in range(0,3):\n",
+    "    for j in range(0,3):\n",
+    "        A[i,j]=round(A[i,j]*10)\n",
+    "    \n",
+    "print 'A = \\n',A\n",
+    "print 'A transpose is:\\n',\n",
+    "Adash=np.transpose(A)\n",
+    "print \"A' = \\n\",Adash\n",
+    "if np.equal(Adash,A).all():\n",
+    "    print \"Since, A' = A, A is a symmetric matrix.\"\n",
+    "else:\n",
+    "    print \"Since, A' is not equal to A, A is not a symmetric matrix.\"\n",
+    "\n",
+    "if np.equal(Adash,-A).all():\n",
+    "    print \"Since, A' = -A, A is a skew-symmetric matrix.\"\n",
+    "else:\n",
+    "    print \"Since, A' is not equal to -A, A is not a skew-symmetric matrix.\"\n",
+    "\n",
+    "A1 = 1./2*(A + Adash)\n",
+    "A2 = 1./2*(A - Adash)\n",
+    "print 'A can be expressed as sum of A1 and A2'\n",
+    "print 'i.e., A = A1 + A2'\n",
+    "print 'A1 = \\n',A1\n",
+    "print 'A2 = \\n',A2\n",
+    "print 'A1 + A2 = \\n',A1 + A2"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter7_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter7_1.ipynb
new file mode 100644
index 00000000..bd958b35
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter7_1.ipynb
@@ -0,0 +1,239 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 7 - The rational and jordan forms"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 239 Example 7.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "Matrix([[5, -6, -6], [-1, 4, 2], [3, -6, -4]])\n",
+      "Characteristic polynomial for linear operator T on R**3 will be:\n",
+      "f =  x**3 - 5*x**2 + 8*x - 4\n",
+      "or\n",
+      "(x-1)(x-2)**2\n",
+      "The minimal polynomial for T is:\n",
+      "p =  (x - 2)*(x - 1)\n",
+      "or\n",
+      "p = (x-1)(x-2)\n",
+      "So in cyclic decomposition of T, a1 will have p as its T-annihilator.\n",
+      "Another vector a2 that generate cyclic subspace of dimension 1 will have its T-annihilator as p2.\n",
+      "p2 =  x - 2\n",
+      "pp2 =  (x - 2)**2*(x - 1)\n",
+      "i.e., pp2 = f\n",
+      "Therefore, A is similar to B\n",
+      "B = \n",
+      "[[ 0 -2  0]\n",
+      " [ 1  3  0]\n",
+      " [ 0  0  2]]\n",
+      "Thus, we can see thet Matrix of T in ordered basis is B\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "A = sp.Matrix(([5, -6, -6],[-1, 4 ,2],[3, -6, -4]))\n",
+    "print 'A = \\n',A\n",
+    "x=sp.Symbol('x')\n",
+    "f = A.charpoly(x).as_expr()\n",
+    "print 'Characteristic polynomial for linear operator T on R**3 will be:'\n",
+    "print 'f = ',f\n",
+    "print 'or'\n",
+    "print '(x-1)(x-2)**2'\n",
+    "print 'The minimal polynomial for T is:'\n",
+    "p = (x-1)*(x-2)#\n",
+    "print 'p = ',p\n",
+    "print 'or'\n",
+    "print 'p = (x-1)(x-2)'\n",
+    "print 'So in cyclic decomposition of T, a1 will have p as its T-annihilator.'\n",
+    "print 'Another vector a2 that generate cyclic subspace of dimension 1 will have its T-annihilator as p2.'\n",
+    "p2 = x-2#\n",
+    "print 'p2 = ',p2\n",
+    "print 'pp2 = ',p*p2\n",
+    "print 'i.e., pp2 = f'\n",
+    "print 'Therefore, A is similar to B'\n",
+    "B = np.array([[0, -2, 0],[1, 3, 0],[0, 0 ,2]])\n",
+    "print 'B = \\n',B\n",
+    "print 'Thus, we can see thet Matrix of T in ordered basis is B'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 247 Example 7.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "2   0   0\n",
+      "a   2   0\n",
+      "b   c   -1\n",
+      "A = \n",
+      "Matrix([[2, 0, 0], [1, 2, 0], [0, 0, -1]])\n",
+      "Characteristic polynomial for A is:\n",
+      "p =  x**3 - 3*x**2 + 4\n",
+      "In this case, minimal polynomial is same as characteristic polynomial.\n",
+      "-----------------------------------------\n",
+      "A = \n",
+      "Matrix([[2, 0, 0], [0, 2, 0], [0, 0, -1]])\n",
+      "Characteristic polynomial for A is:\n",
+      "p =  x**3 - 3*x**2 + 4\n",
+      "In this case, minimal polynomial is: (x-2)(x+1)\n",
+      "or\n",
+      "(x - 2)*(x + 1)\n",
+      "(A-2I)(A+I) = \n",
+      "0    0   0\n",
+      "3a   0   0\n",
+      "ac   0   0\n",
+      "if a = 0, A is similar to diagonal matrix.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "print 'A = '\n",
+    "print '2   0   0'\n",
+    "print 'a   2   0'\n",
+    "print 'b   c   -1'\n",
+    "a = 1#\n",
+    "b = 0#\n",
+    "c = 0#\n",
+    "A = sp.Matrix(([2, 0, 0],[a, 2, 0],[b, c, -1]))\n",
+    "print 'A = \\n',A\n",
+    "print 'Characteristic polynomial for A is:'\n",
+    "x=sp.Symbol('x')\n",
+    "print 'p = ',A.charpoly(x).as_expr()\n",
+    "print 'In this case, minimal polynomial is same as characteristic polynomial.'\n",
+    "print '-----------------------------------------'\n",
+    "a = 0#\n",
+    "b = 0#\n",
+    "c = 0#\n",
+    "A = sp.Matrix(([2, 0, 0],[a, 2, 0],[b, c, -1]))\n",
+    "print 'A = \\n',A\n",
+    "print 'Characteristic polynomial for A is:'\n",
+    "x=sp.Symbol('x')\n",
+    "print 'p = ',A.charpoly(x).as_expr()\n",
+    "print 'In this case, minimal polynomial is:',\n",
+    "print '(x-2)(x+1)'\n",
+    "print 'or'\n",
+    "s = (x-2)*(x+1)#\n",
+    "print s\n",
+    "print '(A-2I)(A+I) = '\n",
+    "print '0    0   0'\n",
+    "print '3a   0   0'\n",
+    "print 'ac   0   0'\n",
+    "print 'if a = 0, A is similar to diagonal matrix.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 247 Example 7.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A = \n",
+      "2   0   0   0\n",
+      "1   2   0   0\n",
+      "0   0   2   0\n",
+      "0   0   a   2\n",
+      "Considering a = 1\n",
+      "Characteristic polynomial for A is:\n",
+      "p =  x**4 - 8*x**3 + 24*x**2 - 32*x + 16\n",
+      "or\n",
+      "(x-2)**4\n",
+      "Minimal polynomial for A =\n",
+      "(x-2)**2\n",
+      "For a = 0 and a = 1, characteristic and minimal polynomial are same.\n",
+      "But for a=0, the solution space of (A - 2I) has 3 dimension whereas for a = 1, it has 2 dimension. \n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "import sympy as sp\n",
+    "print 'A = '\n",
+    "print '2   0   0   0'\n",
+    "print '1   2   0   0'\n",
+    "print '0   0   2   0'\n",
+    "print '0   0   a   2'\n",
+    "print 'Considering a = 1'\n",
+    "A = sp.Matrix(([2, 0 ,0 ,0],[1, 2, 0, 0],[0, 0 ,2 ,0],[0, 0, 1, 2]))\n",
+    "x=sp.Symbol('x')\n",
+    "p = A.charpoly(x).as_expr()\n",
+    "print 'Characteristic polynomial for A is:'\n",
+    "print 'p = ',p\n",
+    "print 'or'\n",
+    "print '(x-2)**4'\n",
+    "print 'Minimal polynomial for A ='\n",
+    "print '(x-2)**2'\n",
+    "print 'For a = 0 and a = 1, characteristic and minimal polynomial are same.'\n",
+    "print 'But for a=0, the solution space of (A - 2I) has 3 dimension whereas for a = 1, it has 2 dimension. '"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter8_1.ipynb b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter8_1.ipynb
new file mode 100644
index 00000000..6f1bfe72
--- /dev/null
+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter8_1.ipynb
@@ -0,0 +1,561 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 8 - Inner product spaces"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 271 Example 8.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "n =  5\n",
+      "a =  [[ 1.  4.  2.  8.  8.]]\n",
+      "b =  [[ 10.   8.   3.   1.   8.]]\n",
+      "Then, (a|b) = \n",
+      "\n",
+      "[[ 10.  40.  20.  80.  80.]\n",
+      " [  8.  32.  16.  64.  64.]\n",
+      " [  3.  12.   6.  24.  24.]\n",
+      " [  1.   4.   2.   8.   8.]\n",
+      " [  8.  32.  16.  64.  64.]]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "n=np.random.randint(2,9)\n",
+    "a=np.random.rand(1,n)\n",
+    "b=np.random.rand(1,n)\n",
+    "for i in range(0,n):\n",
+    "    a[0,i]=round(a[0,i]*10)\n",
+    "    b[0,i]=round(b[0,i]*10)\n",
+    "print 'n = ',n\n",
+    "print 'a = ',a\n",
+    "print 'b = ',b\n",
+    "print 'Then, (a|b) = \\n\\n',a*np.transpose(b)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 271 Example 8.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a =  [[ 4.  3.]]\n",
+      "b =  [[ 7.  5.]]\n",
+      "Then, a|b =  47.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "a=np.random.rand(1,2)\n",
+    "b=np.random.rand(1,2)\n",
+    "for i in range(0,2):\n",
+    "    a[0,i]=round(a[0,i]*10)\n",
+    "    b[0,i]=round(b[0,i]*10)\n",
+    "print 'a = ',a\n",
+    "print 'b = ',b\n",
+    "x1 = a[0,0]#\n",
+    "x2 = a[0,1]#\n",
+    "y1 = b[0,0]#\n",
+    "y2 = b[0,1]#\n",
+    "t = x1*y1 - x2*y1 - x1*y2 + 4*x2*y2#\n",
+    "print 'Then, a|b = ',t"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 307 Example 8.28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x1 and x2  are two real nos. i.e., x1**2 + x2**2 = 1\n",
+      "x1 =  0.383547227589\n",
+      "x2 =  0.923521263539\n",
+      "B = \n",
+      "[[ 0.38354723  0.92352126  0.        ]\n",
+      " [ 0.          1.          0.        ]\n",
+      " [ 0.          0.          1.        ]]\n",
+      "Applying Gram-Schmidt process to B:\n",
+      "a1 =  [ 0.38354723  0.92352126  0.        ]\n",
+      "a2 =  [-0.35421402  0.14710848  0.        ]\n",
+      "a3 =  [0 0 1]\n",
+      "U = \n",
+      "[[[ 0.38354723  0.92352126  0.        ]]\n",
+      "\n",
+      " [[-0.92352126  0.38354723  0.        ]]\n",
+      "\n",
+      " [[ 0.          0.          1.        ]]]\n",
+      "M = \n",
+      "[[ 1.          0.          0.        ]\n",
+      " [-2.40784236  2.60724085  0.        ]\n",
+      " [ 0.          0.          1.        ]]\n",
+      "inverse(M) * U =  [[[  3.83547228e-01  -4.25822963e-17   0.00000000e+00]\n",
+      "  [  3.54214020e-01   3.54214020e-01   0.00000000e+00]\n",
+      "  [  0.00000000e+00   0.00000000e+00   0.00000000e+00]]\n",
+      "\n",
+      " [[ -9.23521264e-01  -1.76848356e-17   0.00000000e+00]\n",
+      "  [ -8.52891524e-01   1.47108476e-01   0.00000000e+00]\n",
+      "  [ -0.00000000e+00   0.00000000e+00   0.00000000e+00]]\n",
+      "\n",
+      " [[  0.00000000e+00  -0.00000000e+00   0.00000000e+00]\n",
+      "  [  0.00000000e+00   0.00000000e+00   0.00000000e+00]\n",
+      "  [  0.00000000e+00   0.00000000e+00   1.00000000e+00]]]\n",
+      "So, B = inverse(M) * U\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "print 'x1 and x2  are two real nos. i.e., x1**2 + x2**2 = 1'\n",
+    "x1 = np.random.rand()\n",
+    "x2 = np.sqrt(1 - x1**2)\n",
+    "print 'x1 = ',x1\n",
+    "print 'x2 = ',x2\n",
+    "B = np.array([[x1, x2, 0],[0, 1, 0],[0, 0, 1]])\n",
+    "print 'B = \\n',B\n",
+    "print 'Applying Gram-Schmidt process to B:'\n",
+    "a1 = np.array([x1, x2, 0])\n",
+    "a2 = np.array([0 ,1 ,0]) - x2 * np.array([x1 ,x2 ,0])\n",
+    "a3 = np.array([0, 0, 1])\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "U = np.array([[a1],[a2/x1],[a3]])\n",
+    "print 'U = \\n',U\n",
+    "M = np.array([[1, 0, 0],[-x2/x1, 1/x1, 0],[0, 0, 1]])\n",
+    "print 'M = \\n',M\n",
+    "print 'inverse(M) * U = ',np.linalg.inv(M) * U\n",
+    "print 'So, B = inverse(M) * U'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 278 Example 8.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(x,y) =  [[ 5.  3.]]\n",
+      "(-y,x) =  [-3.0, 5.0]\n",
+      "Inner product of these vectors is:\n",
+      "(x,y)|(-y,x) =  0.0\n",
+      "So, these are orthogonal.\n",
+      "------------------------------------------\n",
+      "If inner product is defined as:\n",
+      "(x1,x2)|(y1,y2) = x1y1- x2y1 - x1y2 + 4x2y2\n",
+      "Then, (x,y)|(-y,x) = -x*y+y**2-x**2+4*x*y = 0 if,\n",
+      "y = 1/2(-3 + sqrt(13))*x\n",
+      "or\n",
+      "y = 1/2(-3 - sqrt(13))*x\n",
+      "Hence,\n",
+      "[[ 5.  3.]]\n",
+      "is orthogonal to\n",
+      "[-3.0, 5.0]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "#a = round(rand(1,2) * 10)#\n",
+    "a=np.random.rand(1,2)\n",
+    "for j in [0,1]:\n",
+    "    a[0,j]=round(a[0,j]*10)\n",
+    "\n",
+    "x = a[0,0]\n",
+    "y = a[0,1]\n",
+    "b = [-y, x]#\n",
+    "print '(x,y) = ',a\n",
+    "print '(-y,x) = ',b\n",
+    "print 'Inner product of these vectors is:'\n",
+    "t = -x*y + y*x#\n",
+    "print '(x,y)|(-y,x) = ',t\n",
+    "\n",
+    "print 'So, these are orthogonal.'\n",
+    "print '------------------------------------------'\n",
+    "print 'If inner product is defined as:'\n",
+    "print '(x1,x2)|(y1,y2) = x1y1- x2y1 - x1y2 + 4x2y2'\n",
+    "print 'Then, (x,y)|(-y,x) = -x*y+y**2-x**2+4*x*y = 0 if,'\n",
+    "print 'y = 1/2(-3 + sqrt(13))*x'\n",
+    "print 'or'\n",
+    "print 'y = 1/2(-3 - sqrt(13))*x'\n",
+    "print 'Hence,'\n",
+    "if y == (1./2*(-3 + np.sqrt(13))*x) or (1./2*(-3 - np.sqrt(13))*x):\n",
+    "    print a\n",
+    "    print 'is orthogonal to'\n",
+    "    print b\n",
+    "else:\n",
+    "    print a\n",
+    "    print 'is not orthogonal to'\n",
+    "    print b\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 282 Example 8.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "b1 =  [3 0 4]\n",
+      "b2 =  [-1  0  7]\n",
+      "b3 =  [ 2  9 11]\n",
+      "Applying the Gram-Schmidt process to b1,b2,b3:\n",
+      "a1 =  [3 0 4]\n",
+      "a2 =  [2 0 3]\n",
+      "a3 =  [2 9 4]\n",
+      "{a1,a2,a3} are mutually orthogonal and hence forms orthogonal basis for R**3\n",
+      "Any arbitrary vector {x1,x2,x3} in R**3 can be expressed as:\n",
+      "y = {x1,x2,x3} = (3*x1 + 4*x3)/25*a1 + (-4*x1 + 3*x3)/25*a2 + x2/9*a3\n",
+      "x1 =  1\n",
+      "x2 =  2\n",
+      "x3 =  3\n",
+      "y =  [0 0 0]\n",
+      "i.e. y = [x1 x2 x3], according to above equation.\n",
+      "Hence, we get the orthonormal basis as:\n",
+      ", [ 0.6  0.   0.8]\n",
+      ", [ 0.4  0.   0.6]\n",
+      "[ 0.22222222  1.          0.44444444]\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "b1 = np.array([3, 0, 4])\n",
+    "b2 = np.array([-1 ,0 ,7])\n",
+    "b3 = np.array([2 ,9 ,11])\n",
+    "print 'b1 = ',b1\n",
+    "print 'b2 = ',b2\n",
+    "print 'b3 = ',b3\n",
+    "print 'Applying the Gram-Schmidt process to b1,b2,b3:'\n",
+    "a1 = b1\n",
+    "print 'a1 = ',a1\n",
+    "a2 = b2-(np.transpose((b2*np.transpose(b1)))/25*b1)\n",
+    "print 'a2 = ',a2\n",
+    "a3 = b3-(np.transpose(b3*np.transpose(b1))/25*b1) - (np.transpose(b3*np.transpose(a2))/25*a2)\n",
+    "print 'a3 = ',a3\n",
+    "print '{a1,a2,a3} are mutually orthogonal and hence forms orthogonal basis for R**3'\n",
+    "print 'Any arbitrary vector {x1,x2,x3} in R**3 can be expressed as:'\n",
+    "print 'y = {x1,x2,x3} = (3*x1 + 4*x3)/25*a1 + (-4*x1 + 3*x3)/25*a2 + x2/9*a3'\n",
+    "x1 = 1#\n",
+    "x2 = 2#\n",
+    "x3 = 3#\n",
+    "y = (3*x1 + 4*x3)/25*a1 + (-4*x1 + 3*x3)/25*a2 + x2/9*a3#\n",
+    "print 'x1 = ',x1\n",
+    "print 'x2 = ',x2\n",
+    "print 'x3 = ',x3\n",
+    "print 'y = ',y\n",
+    "print 'i.e. y = [x1 x2 x3], according to above equation.'\n",
+    "print 'Hence, we get the orthonormal basis as:'\n",
+    "\n",
+    "print ',',1./5*a1\n",
+    "print ',',1./5*a2\n",
+    "print 1/9.*a3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 283 Example 8.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A =  [[ 1.25598176  1.81258697]\n",
+      " [ 0.6193707   0.80341686]]\n",
+      "b1 =  [ 1.25598176  1.81258697]\n",
+      "b2 =  [ 0.6193707   0.80341686]\n",
+      "Applying the orthogonalization process to b1,b2:\n",
+      "[1.255981755902444, 1.8125869670307564] a1 = \n",
+      "[] a2 = \n",
+      "a2 is not equal to zero if and only if b1 and b2 are linearly independent.\n",
+      "That is, if determinant of A is non-zero.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "A = np.random.rand(2,2)\n",
+    "A[0,:] = A[0,:] + 1# #so b1 is not equal to zero\n",
+    "a = A[0,0]\n",
+    "b = A[0,1]\n",
+    "c = A[1,0]\n",
+    "d = A[1,1]\n",
+    "b1 = A[0,:]\n",
+    "b2 = A[1,:]\n",
+    "print 'A = ',A\n",
+    "print 'b1 = ',b1\n",
+    "print 'b2 = ',b2\n",
+    "print 'Applying the orthogonalization process to b1,b2:'\n",
+    "\n",
+    "a1 = [a, b]\n",
+    "a2 = (np.linalg.det(A)/(a**2 + b**2))*[-np.transpose(b), np.transpose(a)]\n",
+    "print a1,'a1 = '\n",
+    "print a2,'a2 = '\n",
+    "print 'a2 is not equal to zero if and only if b1 and b2 are linearly independent.'\n",
+    "print 'That is, if determinant of A is non-zero.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 286 Example 8.14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "v =  [-10   2   8]\n",
+      "u =  [ 3 12 -1]\n",
+      "Orthogonal projection of v1 on subspace W spanned by v2 is given by:\n",
+      "[-3  0  1]\n",
+      "Orthogonal projection of R**3 on W is the linear transformation E given by:\n",
+      "(x1,x2,x3) -> (3*x1 + 12*x2 - x3)/%d * (3 12 -1) 154\n",
+      "Rank(E) = 1\n",
+      "Nullity(E)  = 2\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "v = np.array([-10 ,2 ,8])\n",
+    "u = np.array([3, 12, -1])\n",
+    "print 'v = ',v\n",
+    "print 'u = ',u\n",
+    "print 'Orthogonal projection of v1 on subspace W spanned by v2 is given by:'\n",
+    "a = (np.transpose(u*np.transpose(v)))/(u[0]**2 + u[1]**2 + u[2]**2) * u\n",
+    "print a\n",
+    "print 'Orthogonal projection of R**3 on W is the linear transformation E given by:'\n",
+    "print '(x1,x2,x3) -> (3*x1 + 12*x2 - x3)/%d * (3 12 -1)',(u[0]**2 + u[1]**2 + u[2]**2)\n",
+    "print 'Rank(E) = 1'\n",
+    "print 'Nullity(E)  = 2'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 288 Example 8.15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "f = (sqrt(2)*cos(2*pi*t) + sqrt(2)*sin(4*pi*t))**2\n",
+      "Integration (f dt) in limits 0 to 1 =  2.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy.mpmath import quad,cos,sin,pi,sqrt\n",
+    "\n",
+    "#part c\n",
+    "print 'f = (sqrt(2)*cos(2*pi*t) + sqrt(2)*sin(4*pi*t))**2'\n",
+    "print 'Integration (f dt) in limits 0 to 1 = ',\n",
+    "x0 = 0#\n",
+    "x1 = 1#\n",
+    "X = quad(lambda t:(sqrt(2)*cos(2*pi*t) + sqrt(2)*sin(4*pi*t))**2,[x0,x1])\n",
+    "print X"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Page 294 Example 8.17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Matrix of projection E in orthonormal basis is:\n",
+      "A = 1/154  *   [[  9  36  -3]\n",
+      " [ 36 144 -12]\n",
+      " [ -3 -12   1]]\n",
+      "A* =  [[  9  36  -3]\n",
+      " [ 36 144 -12]\n",
+      " [ -3 -12   1]]\n",
+      "Since, E = E* and A = A*, then A is also the matrix of E*\n",
+      "a1 =  [154, 0, 0]\n",
+      "a2 =  [145, -36, 3]\n",
+      "a3 =  [-36, 10, 12]\n",
+      "{a1,a2,a3} is the basis.\n",
+      "Ea1 =  [9, 36, -3]\n",
+      "Ea2 =  [0, 0, 0]\n",
+      "Ea3 =  [0, 0, 0]\n",
+      "Matrix B of E in the basis is:\n",
+      "B = \n",
+      "[[-1  0  0]\n",
+      " [-1  0  0]\n",
+      " [ 0  0  0]]\n",
+      "B* = \n",
+      "[[-1 -1  0]\n",
+      " [ 0  0  0]\n",
+      " [ 0  0  0]]\n",
+      "Since, B is not equal to B*, B is not the matrix of E*\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np\n",
+    "#Equation given in example 14 is used.\n",
+    "def transform(x,y,z):\n",
+    "    x1 = 3*x#\n",
+    "    x2 = 12*y#\n",
+    "    x3 = -z#\n",
+    "    m = [x1 ,x2, x3]\n",
+    "    return m\n",
+    "\n",
+    "print 'Matrix of projection E in orthonormal basis is:'\n",
+    "t1 = transform(3,3,3)#\n",
+    "t2 = transform(12,12,12)#\n",
+    "t3 = transform(-1,-1,-1)#\n",
+    "A = np.vstack([t1,t2,t3])#[t1# t2# t3]#\n",
+    "print 'A = 1/154  *  ',A\n",
+    "\n",
+    "A1 = np.transpose(np.conj(A))\n",
+    "print 'A* = ',A1\n",
+    "print 'Since, E = E* and A = A*, then A is also the matrix of E*'\n",
+    "a1 = [154, 0, 0]#\n",
+    "a2 = [145 ,-36, 3]#\n",
+    "a3 = [-36 ,10 ,12]#\n",
+    "print 'a1 = ',a1\n",
+    "print 'a2 = ',a2\n",
+    "print 'a3 = ',a3\n",
+    "print '{a1,a2,a3} is the basis.'\n",
+    "Ea1 = [9 ,36 ,-3]#\n",
+    "Ea2 = [0 ,0, 0]#\n",
+    "Ea3 = [0 ,0 ,0]#\n",
+    "print 'Ea1 = ',Ea1\n",
+    "print 'Ea2 = ',Ea2\n",
+    "print 'Ea3 = ',Ea3\n",
+    "B = np.array([[-1, 0, 0],[-1, 0 ,0],[0, 0, 0]])\n",
+    "print 'Matrix B of E in the basis is:'\n",
+    "print 'B = \\n',B\n",
+    "B1 = np.transpose(np.conj(B))\n",
+    "print 'B* = \\n',B1\n",
+    "print 'Since, B is not equal to B*, B is not the matrix of E*'"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/InverseMatrix_1.png b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/InverseMatrix_1.png
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diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/determinant_1.png b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/determinant_1.png
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diff --git a/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/scalorpolynomial_1.png b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/scalorpolynomial_1.png
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+++ b/Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/scalorpolynomial_1.png