{
"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"
]
}
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