91Ó°ÊÓ

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

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

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

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

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

Conversely, let$\mathrm{σ}$be a singular value of A. Then by definition,${\mathrm{σ}}^2$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 $,{\mathrm{σ}}^2$is an eigenvalue of$R^{-1}{A}^{t}AR.$

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

Thus A and SAR have the same singular values.