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Q32E

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Linear Algebra With Applications
Found in: Page 413
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Consider an matrix A, an orthogonal n×nmatrix S, and an orthogonal m×mmatrix 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 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, σ be a singular value of SAR. Then by definition,σ2 is an eigenvalue of.(SAR)t(SAR) Now we have

(SAR)t(SAR)=RtAtStSAR=RtAtStSAR=RtAtInAR[SinceSisanorthogonalmatrix]=RtAtAR=R-1AtARSinceRisanorthogonalmatrix,soRt=R-1

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

Henceσ2 is an eigen value of R-1AtAR Clearly, the matrixR-1AtAR is similar to the matrix.

Hence R-1AtARand AtAhave same eigen values .Therefore, σ2is an eigen value of Thus is a singular value of A.

Conversely, let σ be a singular value of A. Then by definition,σ2 is an eigenvalue of AtA.Since, the matrix R-1AtAR is similar to the matrix AtA , therefor R-1AtARandAtAhave same eigenvalues. Thus ,σ2is an eigenvalue of R-1AtAR.

Now as, R-1AtAR=(SAR)t(SAR), therefore σ is a singular value of SAR.

Thus A and SAR have the same singular values.

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