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Expert-verified Found in: Page 413 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider an matrix A, an orthogonal ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$matrix S, and an orthogonal ${\mathbit{m}}{\mathbf{×}}{\mathbit{m}}$matrix R. Compare the singular values of A with those of SAR.

A and SAR have the same singular values

See the step by step solution

## Step 1: To find the value of SAR

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

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

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

## Step 2: Compare the singular values of A with those of SAR

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

Thus A and SAR have the same singular values. ### Want to see more solutions like these? 