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Q32E

Expert-verifiedFound in: Page 413

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Consider an**** matrix A, an orthogonal **** ${\mathit{n}}{\mathbf{\times}}{\mathit{n}}$matrix S, and an orthogonal ${\mathit{m}}{\mathbf{\times}}{\mathit{m}}$matrix R****. Compare the singular values of A with those of SAR.**

A and SAR have the same singular values

Given that, A is an matrix, S is an orthogonal $n\times n$ matrix and R is an orthogonal matrix.

Let, $\sigma $ be a singular value of SAR. Then by definition,${\sigma}^{2}$ is an eigenvalue of.$(SAR{)}^{t}\left(SAR\right)$ Now we have

$\begin{array}{rcl}(SAR{)}^{t}\left(SAR\right)& =& {R}^{t}{A}^{t}{S}^{t}SAR={R}^{t}{A}^{t}\left({S}^{t}S\right)AR\\ & =& {R}^{t}{A}^{t}\left({I}_{n}\right)AR[SinceSisanorthogonalmatrix]\\ \u200a& =& {R}^{t}{A}^{t}AR\\ & =& {R}^{-1}{A}^{t}AR\left[SinceRisanorthogonalmatrix,so{R}^{t}={R}^{-1}\right]\\ & & \end{array}$

Hence${\sigma}^{2}$ is an eigen value of ${R}^{-1}{A}^{t}AR$ Clearly, the matrix${R}^{-1}{A}^{t}AR$ is similar to the matrix.

Hence ${R}^{-1}{A}^{t}AR$and ${A}^{t}A$have same eigen values .Therefore, ${\sigma}^{2}$is an eigen value of Thus is a singular value of A.

Conversely, let $\sigma $ be a singular value of A. Then by definition,${\sigma}^{2}$ is an eigenvalue of ${A}^{t}A.$Since, the matrix ${R}^{-1}{A}^{t}AR$ is similar to the matrix ${A}^{t}A$ , therefor ${R}^{-1}{A}^{t}ARand{A}^{t}A$have same eigenvalues. Thus $,{\sigma}^{2}$is an eigenvalue of ${R}^{-1}{A}^{t}AR.$

Now as, ${R}^{-1}{A}^{t}AR=(SAR{)}^{t}\left(SAR\right),$ therefore $\sigma $ is a singular value of SAR.

Thus A and SAR have the same singular values.

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