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 $\Sigma$?
Best Answer
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.