Exercise 1.8.5

Suppose we have $A=U\mathrm{\Sigma }{V}^T$ where $U,V$ are square orthogonal matrices, $V^T=V^{-1}$ and $U^T=U^{-1}$, so we have $A^T=V\mathrm{\Sigma }{U}^T$, the matrix $A^T$ has the same singular values as $A$ but with right singular matrix $U$ and left singular matrix $V$ instead. So we have $\left|\right|A\left|\right|=\left|\right|A^T\left|\right|$ for all matrices because $\left|\right|A\left|\right|=\left|\right|U\mathrm{\Sigma }{V}^T\left|\right|=\left|\right|\mathrm{\Sigma }\left|\right|$

This is not true that $\left|\right|\mathit{Ax}\left|\right|=\left|\right|A^{T}x\left|\right|$ for all vectors, because $\left|\right|\mathit{Ax}|{\text{|}}^2=x^T{A}^T\mathit{Ax}=x^{T}V{\mathrm{\Sigma }}^2{V}^{T}x$ while $\left|\right|A^{T}x|{\text{|}}^2=x^{T}A{A}^{T}x=x^{T}U{\mathrm{\Sigma }}^2{U}^{T}x$.