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ME3255S2017/lecture_13/lecture_13.ipynb

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{
"cells": [
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"%plot --format svg"
]
},
{
"cell_type": "code",
"execution_count": 48,
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"source": [
"setdefaults"
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"source": [
"## My question from last class \n",
"\n",
"\n",
"## Your questions from last class\n",
"\n",
"1. Need more linear algebra review\n",
" \n",
" -We will keep doing Linear Algebra, try the practice problems in 'linear_algebra'\n",
"\n",
"2. How do I do HW3? \n",
" \n",
" -demo today\n",
"\n",
"3. For hw4 is the spring constant (K) suppose to be given? \n",
" \n",
" -yes, its 30 N/m\n",
" \n",
"4. Deapool or Joker?"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Matrix Inverse and Condition\n",
"\n",
"\n",
"Considering the same solution set:\n",
"\n",
"$y=Ax$\n",
"\n",
"If we know that $A^{-1}A=I$, then \n",
"\n",
"$A^{-1}y=A^{-1}Ax=x$\n",
"\n",
"so \n",
"\n",
"$x=A^{-1}y$\n",
"\n",
"Where, $A^{-1}$ is the inverse of matrix $A$.\n",
"\n",
"$2x_{1}+x_{2}=1$\n",
"\n",
"$x_{1}+3x_{2}=1$\n",
"\n",
"$Ax=y$\n",
"\n",
"$\\left[ \\begin{array}{cc}\n",
"2 & 1 \\\\\n",
"1 & 3 \\end{array} \\right]\n",
"\\left[\\begin{array}{c} \n",
"x_{1} \\\\ \n",
"x_{2} \\end{array}\\right]=\n",
"\\left[\\begin{array}{c} \n",
"1 \\\\\n",
"1\\end{array}\\right]$\n",
"\n",
"$A^{-1}=\\frac{1}{2*3-1*1}\\left[ \\begin{array}{cc}\n",
"3 & 1 \\\\\n",
"-1 & 2 \\end{array} \\right]=\n",
"\\left[ \\begin{array}{cc}\n",
"3/5 & -1/5 \\\\\n",
"-1/5 & 2/5 \\end{array} \\right]$\n"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A =\n",
"\n",
" 2 1\n",
" 1 3\n",
"\n",
"invA =\n",
"\n",
" 0.60000 -0.20000\n",
" -0.20000 0.40000\n",
"\n",
"ans =\n",
"\n",
" 1.00000 0.00000\n",
" 0.00000 1.00000\n",
"\n",
"ans =\n",
"\n",
" 1.00000 0.00000\n",
" 0.00000 1.00000\n",
"\n"
]
}
],
"source": [
"A=[2,1;1,3]\n",
"invA=1/5*[3,-1;-1,2]\n",
"\n",
"A*invA\n",
"invA*A"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"How did we know the inverse of A? \n",
"\n",
"for 2$\\times$2 matrices, it is always:\n",
"\n",
"$A=\\left[ \\begin{array}{cc}\n",
"A_{11} & A_{12} \\\\\n",
"A_{21} & A_{22} \\end{array} \\right]$\n",
"\n",
"$A^{-1}=\\frac{1}{det(A)}\\left[ \\begin{array}{cc}\n",
"A_{22} & -A_{12} \\\\\n",
"-A_{21} & A_{11} \\end{array} \\right]$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"$AA^{-1}=\\frac{1}{A_{11}A_{22}-A_{21}A_{12}}\\left[ \\begin{array}{cc}\n",
"A_{11}A_{22}-A_{21}A_{12} & -A_{11}A_{12}+A_{12}A_{11} \\\\\n",
"A_{21}A_{22}-A_{22}A_{21} & -A_{21}A_{12}+A_{22}A_{11} \\end{array} \\right]\n",
"=\\left[ \\begin{array}{cc}\n",
"1 & 0 \\\\\n",
"0 & 1 \\end{array} \\right]$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"What about bigger matrices?\n",
"\n",
"We can use the LU-decomposition\n",
"\n",
"$A=LU$\n",
"\n",
"$A^{-1}=(LU)^{-1}=U^{-1}L^{-1}$\n",
"\n",
"if we divide $A^{-1}$ into n-column vectors, $a_{n}$, then\n",
"\n",
"$Aa_{1}=\\left[\\begin{array}{c} \n",
"1 \\\\ \n",
"0 \\\\ \n",
"\\vdots \\\\\n",
"0 \\end{array} \\right]$\n",
"$Aa_{2}=\\left[\\begin{array}{c} \n",
"0 \\\\ \n",
"1 \\\\ \n",
"\\vdots \\\\\n",
"0 \\end{array} \\right]$\n",
"$Aa_{n}=\\left[\\begin{array}{c} \n",
"0 \\\\ \n",
"0 \\\\ \n",
"\\vdots \\\\\n",
"1 \\end{array} \\right]$\n",
"\n",
"\n",
"Which we can solve for each $a_{n}$ with LU-decomposition, knowing the lower and upper triangular decompositions, then \n",
"\n",
"$A^{-1}=\\left[ \\begin{array}{cccc}\n",
"| & | & & | \\\\\n",
"a_{1} & a_{2} & \\cdots & a_{3} \\\\\n",
"| & | & & | \\end{array} \\right]$\n",
"\n",
"\n",
"$Ld_{1}=\\left[\\begin{array}{c} \n",
"1 \\\\ \n",
"0 \\\\ \n",
"\\vdots \\\\\n",
"0 \\end{array} \\right]$\n",
"$;~Ua_{1}=d_{1}$\n",
"\n",
"$Ld_{2}=\\left[\\begin{array}{c} \n",
"0 \\\\ \n",
"1 \\\\ \n",
"\\vdots \\\\\n",
"0 \\end{array} \\right]$\n",
"$;~Ua_{2}=d_{2}$\n",
"\n",
"$Ld_{n}=\\left[\\begin{array}{c} \n",
"0 \\\\ \n",
"1 \\\\ \n",
"\\vdots \\\\\n",
"n \\end{array} \\right]$\n",
"$;~Ua_{n}=d_{n}$\n",
"\n",
"Consider the following matrix:\n",
"\n",
"$A=\\left[ \\begin{array}{ccc}\n",
"2 & -1 & 0\\\\\n",
"-1 & 2 & -1\\\\\n",
"0 & -1 & 1 \\end{array} \\right]$\n"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A =\n",
"\n",
" 2 -1 0\n",
" -1 2 -1\n",
" 0 -1 1\n",
"\n",
"U =\n",
"\n",
" 2.00000 -1.00000 0.00000\n",
" 0.00000 1.50000 -1.00000\n",
" 0.00000 -1.00000 1.00000\n",
"\n",
"L =\n",
"\n",
" 1.00000 0.00000 0.00000\n",
" -0.50000 1.00000 0.00000\n",
" 0.00000 0.00000 1.00000\n",
"\n"
]
}
],
"source": [
"A=[2,-1,0;-1,2,-1;0,-1,1]\n",
"U=A;\n",
"L=eye(3,3);\n",
"U(2,:)=U(2,:)-U(2,1)/U(1,1)*U(1,:)\n",
"L(2,1)=A(2,1)/A(1,1)"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"L =\n",
"\n",
" 1.00000 0.00000 0.00000\n",
" -0.50000 1.00000 0.00000\n",
" 0.00000 -0.66667 1.00000\n",
"\n",
"U =\n",
"\n",
" 2.00000 -1.00000 0.00000\n",
" 0.00000 1.50000 -1.00000\n",
" 0.00000 0.00000 0.33333\n",
"\n"
]
}
],
"source": [
"L(3,2)=U(3,2)/U(2,2)\n",
"U(3,:)=U(3,:)-U(3,2)/U(2,2)*U(2,:)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Now solve for $d_1$ then $a_1$, $d_2$ then $a_2$, and $d_3$ then $a_{3}$\n",
"\n",
"$Ld_{1}=\\left[\\begin{array}{c} \n",
"1 \\\\ \n",
"0 \\\\ \n",
"\\vdots \\\\\n",
"0 \\end{array} \\right]$\n",
"$;~Ua_{1}=d_{1}$"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"d1 =\n",
"\n",
" 1.00000\n",
" 0.50000\n",
" 0.33333\n",
"\n"
]
}
],
"source": [
"d1=zeros(3,1);\n",
"d1(1)=1;\n",
"d1(2)=0-L(2,1)*d1(1);\n",
"d1(3)=0-L(3,1)*d1(1)-L(3,2)*d1(2)"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a1 =\n",
"\n",
" 1.00000\n",
" 1.00000\n",
" 1.00000\n",
"\n"
]
}
],
"source": [
"a1=zeros(3,1);\n",
"a1(3)=d1(3)/U(3,3);\n",
"a1(2)=1/U(2,2)*(d1(2)-U(2,3)*a1(3));\n",
"a1(1)=1/U(1,1)*(d1(1)-U(1,2)*a1(2)-U(1,3)*a1(3))"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"d2 =\n",
"\n",
" 0.00000\n",
" 1.00000\n",
" 0.66667\n",
"\n"
]
}
],
"source": [
"d2=zeros(3,1);\n",
"d2(1)=0;\n",
"d2(2)=1-L(2,1)*d2(1);\n",
"d2(3)=0-L(3,1)*d2(1)-L(3,2)*d2(2)"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a2 =\n",
"\n",
" 1.0000\n",
" 2.0000\n",
" 2.0000\n",
"\n"
]
}
],
"source": [
"a2=zeros(3,1);\n",
"a2(3)=d2(3)/U(3,3);\n",
"a2(2)=1/U(2,2)*(d2(2)-U(2,3)*a2(3));\n",
"a2(1)=1/U(1,1)*(d2(1)-U(1,2)*a2(2)-U(1,3)*a2(3))"
]
},
{
"cell_type": "code",
"execution_count": 37,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"d3 =\n",
"\n",
" 0\n",
" 0\n",
" 1\n",
"\n"
]
}
],
"source": [
"d3=zeros(3,1);\n",
"d3(1)=0;\n",
"d3(2)=0-L(2,1)*d3(1);\n",
"d3(3)=1-L(3,1)*d3(1)-L(3,2)*d3(2)"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a3 =\n",
"\n",
" 1.00000\n",
" 2.00000\n",
" 3.00000\n",
"\n"
]
}
],
"source": [
"a3=zeros(3,1);\n",
"a3(3)=d3(3)/U(3,3);\n",
"a3(2)=1/U(2,2)*(d3(2)-U(2,3)*a3(3));\n",
"a3(1)=1/U(1,1)*(d3(1)-U(1,2)*a3(2)-U(1,3)*a3(3))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Final solution for $A^{-1}$ is $[a_{1}~a_{2}~a_{3}]$"
]
},
{
"cell_type": "code",
"execution_count": 40,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"invA =\n",
"\n",
" 1.00000 1.00000 1.00000\n",
" 1.00000 2.00000 2.00000\n",
" 1.00000 2.00000 3.00000\n",
"\n",
"ans =\n",
"\n",
" 1.00000 0.00000 0.00000\n",
" 0.00000 1.00000 -0.00000\n",
" -0.00000 -0.00000 1.00000\n",
"\n"
]
}
],
"source": [
"invA=[a1,a2,a3]\n",
"A*invA"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Now the solution of $x$ to $Ax=y$ is $x=A^{-1}y$"
]
},
{
"cell_type": "code",
"execution_count": 44,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"y =\n",
"\n",
" 1\n",
" 2\n",
" 3\n",
"\n",
"x =\n",
"\n",
" 6.0000\n",
" 11.0000\n",
" 14.0000\n",
"\n",
"xbs =\n",
"\n",
" 6.0000\n",
" 11.0000\n",
" 14.0000\n",
"\n",
"ans =\n",
"\n",
" -3.5527e-15\n",
" -8.8818e-15\n",
" -1.0658e-14\n",
"\n",
"ans = 2.2204e-16\n"
]
}
],
"source": [
"y=[1;2;3]\n",
"x=invA*y\n",
"xbs=A\\y\n",
"x-xbs\n",
"eps"
]
},
{
"cell_type": "code",
"execution_count": 61,
"metadata": {
"collapsed": false
},
"outputs": [
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"N=100;\n",
"n=[1:N];\n",
"t_inv=zeros(N,1);\n",
"t_bs=zeros(N,1);\n",
"t_mult=zeros(N,1);\n",
"for i=1:N\n",
" A=rand(i,i);\n",
" tic\n",
" invA=inv(A);\n",
" t_inv(i)=toc;\n",
" b=rand(i,1);\n",
" tic;\n",
" x=A\\b;\n",
" t_bs(i)=toc;\n",
" tic;\n",
" x=invA*b;\n",
" t_mult(i)=toc;\n",
"end\n",
"plot(n,t_inv,n,t_bs,n,t_mult)\n",
"axis([0 100 0 0.002])\n",
"legend('inversion','backslash','multiplication','Location','NorthWest')"
]
},
{
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"source": [
"## Condition of a matrix \n",
"### *just checked in to see what condition my condition was in*\n",
"### Matrix norms\n",
"\n",
"The Euclidean norm of a vector is measure of the magnitude (in 3D this would be: $|x|=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}$) in general t"
]
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"metadata": {
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{
"name": "stdout",
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"text": [
"A =\n",
"\n",
" 2 1\n",
" 1 3\n",
"\n",
"L =\n",
"\n",
" 1.00000 0.00000\n",
" 0.50000 1.00000\n",
"\n",
"U =\n",
"\n",
" 2.00000 1.00000\n",
" 0.00000 2.50000\n",
"\n",
"ans =\n",
"\n",
" 2 1\n",
" 1 3\n",
"\n",
"d =\n",
"\n",
" 1.00000\n",
" 0.50000\n",
"\n",
"y =\n",
"\n",
" 1\n",
" 1\n",
"\n"
]
}
],
"source": [
"A=[2,1;1,3]\n",
"L=[1,0;0.5,1]\n",
"U=[2,1;0,2.5]\n",
"L*U\n",
"\n",
"d=[1;0.5]\n",
"y=L*d"
]
},
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"cell_type": "code",
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{
"name": "stdout",
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"text": [
"ans = 5.0000\n",
"ans = 1\n",
"ans = 5\n",
"ans = 5\n",
"ans = 5.0000\n"
]
}
],
"source": [
"% what is the determinant of L, U and A?\n",
"\n",
"det(A)\n",
"det(L)\n",
"det(U)\n",
"det(L)*det(U)\n",
"det(L*U)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Pivoting for LU factorization\n",
"\n",
"LU factorization uses the same method as Gauss elimination so it is also necessary to perform partial pivoting when creating the lower and upper triangular matrices. \n",
"\n",
"Matlab and Octave use pivoting in the command \n",
"\n",
"`[L,U,P]=lu(A)`\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": 4,
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{
"name": "stdout",
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"text": [
"'lu' is a built-in function from the file libinterp/corefcn/lu.cc\n",
"\n",
" -- Built-in Function: [L, U] = lu (A)\n",
" -- Built-in Function: [L, U, P] = lu (A)\n",
" -- Built-in Function: [L, U, P, Q] = lu (S)\n",
" -- Built-in Function: [L, U, P, Q, R] = lu (S)\n",
" -- Built-in Function: [...] = lu (S, THRES)\n",
" -- Built-in Function: Y = lu (...)\n",
" -- Built-in Function: [...] = lu (..., \"vector\")\n",
" Compute the LU decomposition of A.\n",
"\n",
" If A is full subroutines from LAPACK are used and if A is sparse\n",
" then UMFPACK is used.\n",
"\n",
" The result is returned in a permuted form, according to the\n",
" optional return value P. For example, given the matrix 'a = [1, 2;\n",
" 3, 4]',\n",
"\n",
" [l, u, p] = lu (A)\n",
"\n",
" returns\n",
"\n",
" l =\n",
"\n",
" 1.00000 0.00000\n",
" 0.33333 1.00000\n",
"\n",
" u =\n",
"\n",
" 3.00000 4.00000\n",
" 0.00000 0.66667\n",
"\n",
" p =\n",
"\n",
" 0 1\n",
" 1 0\n",
"\n",
" The matrix is not required to be square.\n",
"\n",
" When called with two or three output arguments and a spare input\n",
" matrix, 'lu' does not attempt to perform sparsity preserving column\n",
" permutations. Called with a fourth output argument, the sparsity\n",
" preserving column transformation Q is returned, such that 'P * A *\n",
" Q = L * U'.\n",
"\n",
" Called with a fifth output argument and a sparse input matrix, 'lu'\n",
" attempts to use a scaling factor R on the input matrix such that 'P\n",
" * (R \\ A) * Q = L * U'. This typically leads to a sparser and more\n",
" stable factorization.\n",
"\n",
" An additional input argument THRES, that defines the pivoting\n",
" threshold can be given. THRES can be a scalar, in which case it\n",
" defines the UMFPACK pivoting tolerance for both symmetric and\n",
" unsymmetric cases. If THRES is a 2-element vector, then the first\n",
" element defines the pivoting tolerance for the unsymmetric UMFPACK\n",
" pivoting strategy and the second for the symmetric strategy. By\n",
" default, the values defined by 'spparms' are used ([0.1, 0.001]).\n",
"\n",
" Given the string argument \"vector\", 'lu' returns the values of P\n",
" and Q as vector values, such that for full matrix, 'A (P,:) = L *\n",
" U', and 'R(P,:) * A (:, Q) = L * U'.\n",
"\n",
" With two output arguments, returns the permuted forms of the upper\n",
" and lower triangular matrices, such that 'A = L * U'. With one\n",
" output argument Y, then the matrix returned by the LAPACK routines\n",
" is returned. If the input matrix is sparse then the matrix L is\n",
" embedded into U to give a return value similar to the full case.\n",
" For both full and sparse matrices, 'lu' loses the permutation\n",
" information.\n",
"\n",
" See also: luupdate, ilu, chol, hess, qr, qz, schur, svd.\n",
"\n",
"Additional help for built-in functions and operators is\n",
"available in the online version of the manual. Use the command\n",
"'doc <topic>' to search the manual index.\n",
"\n",
"Help and information about Octave is also available on the WWW\n",
"at http://www.octave.org and via the help@octave.org\n",
"mailing list.\n"
]
}
],
"source": [
"help lu"
]
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],
"source": [
"% time LU solution vs backslash\n",
"t_lu=zeros(100,1);\n",
"t_bs=zeros(100,1);\n",
"for N=1:100\n",
" A=rand(N,N);\n",
" y=rand(N,1);\n",
" [L,U,P]=lu(A);\n",
"\n",
" tic; d=inv(L)*y; x=inv(U)*d; t_lu(N)=toc;\n",
"\n",
" tic; x=inv(A)*y; t_bs(N)=toc;\n",
"end\n",
"plot([1:100],t_lu,[1:100],t_bs) \n",
"legend('LU decomp','Octave \\\\')"
]
},
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"cell_type": "markdown",
"metadata": {
"collapsed": true
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"source": [
"Consider the problem again from the intro to Linear Algebra, 4 masses are connected in series to 4 springs with K=10 N/m. What are the final positions of the masses? \n",
"\n",
"![Springs-masses](../lecture_09/mass_springs.svg)\n",
"\n",
"The masses haves the following amounts, 1, 2, 3, and 4 kg for masses 1-4. Using a FBD for each mass:\n",
"\n",
"$m_{1}g+k(x_{2}-x_{1})-kx_{1}=0$\n",
"\n",
"$m_{2}g+k(x_{3}-x_{2})-k(x_{2}-x_{1})=0$\n",
"\n",
"$m_{3}g+k(x_{4}-x_{3})-k(x_{3}-x_{2})=0$\n",
"\n",
"$m_{4}g-k(x_{4}-x_{3})=0$\n",
"\n",
"in matrix form:\n",
"\n",
"$\\left[ \\begin{array}{cccc}\n",
"2k & -k & 0 & 0 \\\\\n",
"-k & 2k & -k & 0 \\\\\n",
"0 & -k & 2k & -k \\\\\n",
"0 & 0 & -k & k \\end{array} \\right]\n",
"\\left[ \\begin{array}{c}\n",
"x_{1} \\\\\n",
"x_{2} \\\\\n",
"x_{3} \\\\\n",
"x_{4} \\end{array} \\right]=\n",
"\\left[ \\begin{array}{c}\n",
"m_{1}g \\\\\n",
"m_{2}g \\\\\n",
"m_{3}g \\\\\n",
"m_{4}g \\end{array} \\right]$"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"K =\n",
"\n",
" 20 -10 0 0\n",
" -10 20 -10 0\n",
" 0 -10 20 -10\n",
" 0 0 -10 10\n",
"\n",
"y =\n",
"\n",
" 9.8100\n",
" 19.6200\n",
" 29.4300\n",
" 39.2400\n",
"\n"
]
}
],
"source": [
"k=10; % N/m\n",
"m1=1; % kg\n",
"m2=2;\n",
"m3=3;\n",
"m4=4;\n",
"g=9.81; % m/s^2\n",
"K=[2*k -k 0 0; -k 2*k -k 0; 0 -k 2*k -k; 0 0 -k k]\n",
"y=[m1*g;m2*g;m3*g;m4*g]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"This matrix, K, is symmetric. \n",
"\n",
"`K(i,j)==K(j,i)`\n",
"\n",
"Now we can use,\n",
"\n",
"## Cholesky Factorization\n",
"\n",
"We can decompose the matrix, K into two matrices, $U$ and $U^{T}$, where\n",
"\n",
"$K=U^{T}U$\n",
"\n",
"each of the components of U can be calculated with the following equations:\n",
"\n",
"$u_{ii}=\\sqrt{a_{ii}-\\sum_{k=1}^{i-1}u_{ki}^{2}}$\n",
"\n",
"$u_{ij}=\\frac{a_{ij}-\\sum_{k=1}^{i-1}u_{ki}u_{kj}}{u_{ii}}$\n",
"\n",
"so for K"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"K =\n",
"\n",
" 20 -10 0 0\n",
" -10 20 -10 0\n",
" 0 -10 20 -10\n",
" 0 0 -10 10\n",
"\n"
]
}
],
"source": [
"K"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"u11 = 4.4721\n",
"u12 = -2.2361\n",
"u13 = 0\n",
"u14 = 0\n",
"u22 = 3.8730\n",
"u23 = -2.5820\n",
"u24 = 0\n",
"u33 = 3.6515\n",
"u34 = -2.7386\n",
"u44 = 1.5811\n",
"U =\n",
"\n",
" 4.47214 -2.23607 0.00000 0.00000\n",
" 0.00000 3.87298 -2.58199 0.00000\n",
" 0.00000 0.00000 3.65148 -2.73861\n",
" 0.00000 0.00000 0.00000 1.58114\n",
"\n"
]
}
],
"source": [
"u11=sqrt(K(1,1))\n",
"u12=(K(1,2))/u11\n",
"u13=(K(1,3))/u11\n",
"u14=(K(1,4))/u11\n",
"u22=sqrt(K(2,2)-u12^2)\n",
"u23=(K(2,3)-u12*u13)/u22\n",
"u24=(K(2,4)-u12*u14)/u22\n",
"u33=sqrt(K(3,3)-u13^2-u23^2)\n",
"u34=(K(3,4)-u13*u14-u23*u24)/u33\n",
"u44=sqrt(K(4,4)-u14^2-u24^2-u34^2)\n",
"U=[u11,u12,u13,u14;0,u22,u23,u24;0,0,u33,u34;0,0,0,u44]"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"ans =\n",
"\n",
" 1 1 1 1\n",
" 1 1 1 1\n",
" 1 1 1 1\n",
" 1 1 1 1\n",
"\n"
]
}
],
"source": [
"(U'*U)'==U'*U"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"average time spent for Cholesky factored solution = 1.465964e-05+/-9.806001e-06\n",
"average time spent for backslash solution = 1.555967e-05+/-1.048561e-05\n"
]
}
],
"source": [
"% time solution for Cholesky vs backslash\n",
"t_chol=zeros(1000,1);\n",
"t_bs=zeros(1000,1);\n",
"for i=1:1000\n",
" tic; d=U'*y; x=U\\d; t_chol(i)=toc;\n",
" tic; x=K\\y; t_bs(i)=toc;\n",
"end\n",
"fprintf('average time spent for Cholesky factored solution = %e+/-%e',mean(t_chol),std(t_chol))\n",
"\n",
"fprintf('average time spent for backslash solution = %e+/-%e',mean(t_bs),std(t_bs))"
]
},
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