Singular value decomposition: rotation Suppose that $A \in \mathbb{C}^{m \times n}$ and $B$ ($ \in \mathbb{C}^{n \times m}$) is the matrix obtained by rotating $A$ ninety degrees clockwise. Do $A$ and $B$ have the same singular values? 
My first attempt was to find certain transformation matrices to obtain the 'rotated' matrices of the singular value decomposition of $A$ (=$U \Sigma V^{*}$). So in this way, $U$ and $V$ stay unitary matrices but I don't know what happens with the matrix $\Sigma$? 
 A: Note that the matrix $B$ is a result of transposing $A$ and than permuting the columns. Now apply these operations to the SVD of the matrix $A$: transposing doesn't change singular values, since
$$
(U \Sigma V^*)^T = V^{*T} \Sigma^T U^T
$$
and permuting columns is multiplication by some non-singular square matrix from the left, which doesn't change the "main" part of the SVD decomposition - matrix $\Sigma^T$.
So the singular values will remain the same.
A: Let
$$
A=\begin{bmatrix}a_{11}&...&a_{1n}\\\vdots&\ddots&\vdots\\
a_{m1}&...&a_{mn}\end{bmatrix}_{m\times n}.$$
When we rotate $A$ ninety degrees clockwise we obtain
$$
B=\begin{bmatrix}a_{1m}&...&a_{11}\\\vdots&\ddots&\vdots\\
a_{mn}&...&a_{1n}\end{bmatrix}_{n\times m}.
$$
Note that $B=A^TE$ where
$$
E=\begin{bmatrix}0&...&1\\\vdots&.^{.^.}&\vdots\\
1&...&0\end{bmatrix}_{m\times m}.
$$
Let $A=U_A\Sigma_AV^*_{A}$ be the singular value decomposition of $A$. Then
$$
A^T=\overline{V_A}\Sigma_A^TU_A^T\implies B=A^TE=\overline{V_A}\Sigma_A^TU_A^TE.
$$
So we obtain the singular value decomposition of B as $B=U_B\Sigma_BV^*_{B}$ where
$$
U_B=\overline{V_A},~\Sigma_B=\Sigma_A^T,~V_B=E\overline{U_A}.
$$
Therefore the singular values won't change by rotating $A$ ninety degrees clockwise.
Note: $\overline{A}$ denotes the conjugate matrix of $A$.
