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Mathematics-Online course: Linear Algebra - Normal Forms - Singular Value Decomposition | ||
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>> h=[834, 561, 207, 9]'; >> d=[276, 631, 822, 356, 549]';where .
Let
SV-decomposition of matrix A:
>> A=[1 -1 0 0 ; 1 0 -1 0 ; 1 0 0 -1 ; 0 1 -1 0 ; 0 1 0 -1]; >> [U,S,V]=svd(A) U = 0.5774 0.4082 -0.0000 -0.7071 -0.0000 0.5774 -0.2041 -0.5000 0.3536 0.5000 0.5774 -0.2041 0.5000 0.3536 -0.5000 -0.0000 -0.6124 -0.5000 -0.3536 -0.5000 -0.0000 -0.6124 0.5000 -0.3536 0.5000 S = 2.0000 0 0 0 0 2.0000 0 0 0 0 1.4142 0 0 0 0 0.0000 0 0 0 0 V = 0.8660 0 0 -0.5000 -0.2887 -0.8165 0.0000 -0.5000 -0.2887 0.4082 0.7071 -0.5000 -0.2887 0.4082 -0.7071 -0.5000
Computation of the pseudo inverse by means of the -matrix the entries of which are
>> SP=diag(1./diag(S(1:3,1:3))); >> SP(4,5)=0 SP = 0.5000 0 0 0 0 0 0.5000 0 0 0 0 0 0.7071 0 0 0 0 0 0 0 >> AP=V*SP*U' AP = 0.2500 0.2500 0.2500 -0.0000 -0.0000 -0.2500 0.0000 0.0000 0.2500 0.2500 -0.0000 -0.3750 0.1250 -0.3750 0.1250 0.0000 0.1250 -0.3750 0.1250 -0.3750
Solving the approximation problem by multiplying the right hand side by the pseudo inverse :
>> x=AP*(d-A*h) x = 1.0000 -1.0000 -3.0000 3.0000
Remark: The pseudo inverse can directly be computed by the MATLAB-command pinv.
automatically generated 4/21/2005 |