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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter2 Vector Spaces"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.1.1 Pg: 70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Consider all vectors in R**2 whose components are positive or zero\n",
+ "The subset is first Quadrant of x-y plane,the co-ordinates satisfy x>=0 and y>=0.It is not a subspace.\n",
+ "If the Vector= [1, 1]\n",
+ "Taking a scalar,c=-1\n",
+ "c*v= [-1, -1]\n",
+ "It lies in third Quadrant instead of first,Hence violating the rule(ii).\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Consider all vectors in R**2 whose components are positive or zero'\n",
+ "print 'The subset is first Quadrant of x-y plane,the co-ordinates satisfy x>=0 and y>=0.It is not a subspace.'\n",
+ "v=[1,1]\n",
+ "print 'If the Vector=',v\n",
+ "print 'Taking a scalar,c=-1'\n",
+ "c=-1# #scalar\n",
+ "print 'c*v=',[c*vv for vv in v]\n",
+ "print 'It lies in third Quadrant instead of first,Hence violating the rule(ii).'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.3.2 Pg: 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Given matrix:\n",
+ "[[ 1 3 3 2]\n",
+ " [ 2 6 9 5]\n",
+ " [-1 -3 3 0]]\n",
+ "C2->C2-3*C1\n",
+ "[[ 1 0 3 2]\n",
+ " [ 2 0 9 5]\n",
+ " [-1 0 3 0]]\n",
+ "Here,C2=3*C1,Therefore the columns are linearly dependent.\n",
+ "R3->R3-2*R2+5*R1\n",
+ "[[1 0 3 2]\n",
+ " [2 0 9 5]\n",
+ " [0 0 0 0]]\n",
+ "Here R3=R3-2*R2+5*R1,therefore the rows are linearly dependent.\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import mat\n",
+ "A=mat([[1, 3 ,3 ,2],[2, 6, 9, 5],[-1, -3, 3, 0]])\n",
+ "print 'Given matrix:'\n",
+ "print A\n",
+ "B=A\n",
+ "print 'C2->C2-3*C1'\n",
+ "A[:,1]=A[:,1]-3*A[:,0]\n",
+ "print A\n",
+ "print 'Here,C2=3*C1,Therefore the columns are linearly dependent.'\n",
+ "print 'R3->R3-2*R2+5*R1'\n",
+ "B[2,:]=B[2,:]-2*B[1,:]+5*B[0,:]\n",
+ "print B\n",
+ "print 'Here R3=R3-2*R2+5*R1,therefore the rows are linearly dependent.'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.3.3 Pg: "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A=\n",
+ "[[3 4 2]\n",
+ " [0 1 5]\n",
+ " [0 0 2]]\n",
+ "The columns of the triangular matrix are linearly independent,it has no zeros on the diagonal\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import mat\n",
+ "A=mat([[3, 4 ,2],[0, 1 ,5],[0, 0, 2]])\n",
+ "print 'A=\\n',A\n",
+ "print 'The columns of the triangular matrix are linearly independent,it has no zeros on the diagonal'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.3.4 Pg: 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The columns of the nxn identity matrix are independent.\n",
+ "I=\n",
+ "[[ 1. 0. 0. 0.]\n",
+ " [ 0. 1. 0. 0.]\n",
+ " [ 0. 0. 1. 0.]\n",
+ " [ 0. 0. 0. 1.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import eye\n",
+ "print 'The columns of the nxn identity matrix are independent.'\n",
+ "n=4 # size for identity matrix\n",
+ "I=eye(n)\n",
+ "print 'I=\\n',I"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.3.5 Pg: 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Three columns in R2 cannot be independent.\n",
+ "Given matrix:\n",
+ "[[1 2 1]\n",
+ " [1 2 3]]\n",
+ "U=\n",
+ "[[ 1. 2. 1.]\n",
+ " [ 0. 0. 2.]]\n",
+ "If c3 is 1 ,then back-substitution Uc=0 gives c2=-1,c1=1,With these three weights,the first column minus the second plus the third equals zero ,therefore linearly dependent.\n"
+ ]
+ }
+ ],
+ "source": [
+ "from scipy.linalg import lu\n",
+ "from numpy import mat\n",
+ "print 'Three columns in R2 cannot be independent.'\n",
+ "A=mat([[1, 2, 1],[1, 2, 3]])\n",
+ "print 'Given matrix:\\n',A\n",
+ "L=lu(A)[1]\n",
+ "U=lu(A)[2]\n",
+ "print 'U=\\n',U\n",
+ "print 'If c3 is 1 ,then back-substitution Uc=0 gives c2=-1,c1=1,With these three weights,the first column minus the second plus the third equals zero ,therefore linearly dependent.'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.3.9 Pg: 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "These four columns span the column space U,but they are not independent.\n",
+ "U=\n",
+ "[[1 3 3 2]\n",
+ " [0 0 3 1]\n",
+ " [0 0 0 0]]\n",
+ "The columns that contains pivots (here 1st & 3rd) are a basis for the column space. These columns are independent, and it is easy to see that they span the space.In fact,the column space of U is just the x-y plane withinn R3. C(U) is not the same as the column space C(A) before elimination-but the number of independent columns did not change.\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import mat\n",
+ "print 'These four columns span the column space U,but they are not independent.'\n",
+ "U=mat([[1, 3 ,3, 2],[0, 0 ,3 ,1],[0, 0, 0, 0]])\n",
+ "print 'U=\\n',U\n",
+ "print 'The columns that contains pivots (here 1st & 3rd) are a basis for the column space. These columns are independent, and it is easy to see that they span the space.In fact,the column space of U is just the x-y plane withinn R3. C(U) is not the same as the column space C(A) before elimination-but the number of independent columns did not change.'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.4.1 Pg: 107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A=\n",
+ "[[1 2]\n",
+ " [3 6]]\n",
+ "m= 2\n",
+ "n= 2\n",
+ "rank= 1\n",
+ "Column space= [[1]\n",
+ " [3]]\n",
+ "Null space=\n",
+ "[[-0.89442719]\n",
+ " [ 0.4472136 ]]\n",
+ "Row space=\n",
+ "[[1]\n",
+ " [2]]\n",
+ "Left null sapce=\n",
+ "[[-0.9486833 ]\n",
+ " [ 0.31622777]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import mat,shape\n",
+ "from sympy import Matrix\n",
+ "from scipy import linalg, matrix, compress,transpose\n",
+ "A=mat([[1 ,2],[3, 6]])\n",
+ "print 'A=\\n',A\n",
+ "m=shape(A)[0]\n",
+ "n=shape(A)[1]\n",
+ "print 'm=',m\n",
+ "print 'n=',n\n",
+ "v=Matrix(A).rref()[0]\n",
+ "pivot=Matrix(A).rref()[1]\n",
+ "r=len(pivot)\n",
+ "print 'rank=',r\n",
+ "cs=A[:,r-1]\n",
+ "print 'Column space=',cs\n",
+ "\n",
+ "\n",
+ "def kernel(A, eps=1e-15):\n",
+ " u, s, vh = linalg.svd(A)\n",
+ " null_mask = (s <= eps)\n",
+ " null_space = compress(null_mask, vh, axis=0)\n",
+ " return transpose(null_space)\n",
+ "A=mat([[1 ,2],[3, 6]])\n",
+ "\n",
+ "ns=kernel(A)\n",
+ "print 'Null space=\\n',ns\n",
+ "v=mat(v)\n",
+ "rs=transpose(v[range(0,r),:])\n",
+ "print 'Row space=\\n',rs\n",
+ "lns=kernel(transpose(A))\n",
+ "print 'Left null sapce=\\n',lns"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.4.2 Pg: 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A=\n",
+ "[[4 0 0]\n",
+ " [0 5 0]]\n",
+ "m= 2\n",
+ "n= 3\n",
+ "rank= 2\n",
+ "since m=r=2 ,there exists a right inverse .\n",
+ "Best right inverse=\n",
+ "[[ 0.25 0. ]\n",
+ " [ 0. 0.2 ]\n",
+ " [ 0. 0. ]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import mat,shape,rank,transpose\n",
+ "A=mat([[4, 0, 0],[0, 5, 0]])\n",
+ "print 'A=\\n',A\n",
+ "m=shape(A)[0]\n",
+ "n=shape(A)[1]\n",
+ "print 'm=',m\n",
+ "print 'n=',n\n",
+ "r=rank(A)\n",
+ "print 'rank=',r\n",
+ "print 'since m=r=2 ,there exists a right inverse .'\n",
+ "C=transpose(A)*(A*transpose(A))**-1\n",
+ "print 'Best right inverse=\\n',C"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex:2.5.1 Pg: 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Applying current law Ay=f at nodes 1,2,3:\n",
+ "A' = \n",
+ "[[-1 0 -1 0 -1]\n",
+ " [ 1 -1 0 0 0]\n",
+ " [ 0 1 1 -1 0]]\n",
+ "[[ 0. ]\n",
+ " [ 0. ]\n",
+ " [ 0.59010697]\n",
+ " [ 0. ]\n",
+ " [ 0. ]\n",
+ " [ 0. ]\n",
+ " [ 0.55757785]\n",
+ " [ 0. ]]\n",
+ "The other equation is inv(C)y+Ax=b.The block form of the two equations is:\n",
+ "[[ 1.37051236 0. 0. 0. 0. -1.\n",
+ " 1. 0. ]\n",
+ " [ 0. 1.54412147 0. 0. 0. 0.\n",
+ " -1. 1. ]\n",
+ " [ 0. 0. 9.1949881 0. 0. -1.\n",
+ " 0. 1. ]\n",
+ " [ 0. 0. 0. 13.91590014 0. 0.\n",
+ " 0. -1. ]\n",
+ " [ 0. 0. 0. 0. 2.46587381 -1.\n",
+ " 0. 0. ]\n",
+ " [ -1. 0. -1. 0. -1. 0.\n",
+ " 0. 0. ]\n",
+ " [ 1. -1. 0. 0. 0. 0.\n",
+ " 0. 0. ]\n",
+ " [ 0. 1. 1. -1. 0. 0.\n",
+ " 0. 0. ]]\n",
+ "X=\n",
+ "[['y1']\n",
+ " ['y2']\n",
+ " ['y3']\n",
+ " ['y4']\n",
+ " ['y5']\n",
+ " ['x1']\n",
+ " ['x2']\n",
+ " ['x3']]\n",
+ "X= [[ 0.3717052 ]\n",
+ " [-0.18587265]\n",
+ " [ 0.08836592]\n",
+ " [-0.09750673]\n",
+ " [-0.46007112]\n",
+ " [-1.13447733]\n",
+ " [-1.6439039 ]\n",
+ " [-1.35689395]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "from numpy import mat,transpose,diag,zeros,random,vstack,hstack,linalg\n",
+ "print 'Applying current law A''y=f at nodes 1,2,3:'\n",
+ "A=mat([[-1, 1 ,0],[0, -1, 1],[ -1, 0 ,1],[0, 0 ,-1],[-1, 0, 0]])\n",
+ "print \"A' = \\n\",transpose(A)\n",
+ "C=diag(random.rand(5))# #Taking some values for the resistances.\\\n",
+ "b=zeros([5,1])\n",
+ "b[2,0]=random.rand(1)[0]##Taking some value of the battery.\n",
+ "f=zeros([3,1])\n",
+ "f[1,0]=random.rand(1)[0]##Taking some value of the current source.\n",
+ "B=vstack([b,f])#[b]#f]\n",
+ "print B\n",
+ "print 'The other equation is inv(C)y+Ax=b.The block form of the two equations is:'\n",
+ "#C=[C**-1 A],[np.transpose(A),np.zeros([3,3])]\n",
+ "C1=hstack([linalg.inv(C),A])\n",
+ "C2=hstack([transpose(A),zeros([3,3])])\n",
+ "C=vstack([C1,C2])\n",
+ "print C\n",
+ "X=mat([['y1'],['y2'],['y3'],['y4'],['y5'],['x1'],['x2'],['x3']])\n",
+ "print \"X=\\n\",X\n",
+ "X=linalg.solve(C,B)\n",
+ "print 'X=',X"
+ ]
+ }
+ ],
+ "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
+}
generated by cgit (git 2.34.1) at 2025-04-03 15:35:34 +0000