WeBWork Linear Algebra - Section 6.5 Singular Value Decomposition - Section 6.5 Singluar Value Decomposition Find the singular values of sigma1>=sigma2

# WeBWork Linear Algebra - Section 6.5 Singular Value Decomposition

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Section 6.5 Singluar Value Decomposition Find the singular values of sigma1>=sigma2 of A=[-2,-2,,2,-1,,1,1] ATA=[-2,2,1,,-2,-1,1][-2,-2,,2,-1,,1,1]=[9,3,,3,6] Let det(ATA-alphaI2)=0 where I2 is the 2x2 identity matrix. We have (9-alpha)(6-alpha)-3^2=0 or alpha^2-15alpha+45=0 which has two roots alpha1=(15+3sqrt(5))/2, alpha2=(15-3sqrt(5))/2 Thiat is ATA has two eigenvalues A1,A2 Thus A has singular values sigma1=sqrt(alpha1)=sqrt((15+3sqrt(5)/2), sigma2=sqrt(alpha2)=sqrt((15-3sqrt(5)/2) Find the singular values of sigma1>=sigma2>=sigma3 of A=[-1,0,-4,,4,0,-1] Singular values of a matrix A are the squareroots of eigenvalues of the matrix $$A^TA$$ Now, $$A^TA=\begin{pmatrix}-1&4\\0&0\\-4&-1\end{pmatrix}\begin{pmatrix}-1&0&- 4\\4&0&-1\end{pmatrix}=\begin{pmatrix}17&0&0\\0&0&0\\0&0&17\end{pmatrix}$$ since the matrix formed is already diagonal, so the eigenvalues of $$A^TA$$ are 17 and 0 thus, the singular values of A are $$s1=\sqrt{17}, s2=-\sqrt{17}, s3=0$$ Let A=[5.5,-1;2.5,-5;-2.5,5;-5.5,1]. A singular value decomposition of A is as follows: A=[-0.5,0.5,-0.5,0.5;-0.5,-0.5,0.5,0.5;0.5,0.5,0.5,0.5;0.5,-0.5,-0.5,0.5][10,0;0,5,0,0,0,0][- 0.8,0.6;0.6,0.8] (A) Find the closest (with respect to the Frobenius norm) matrix of rank 1 to A.

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