SVD and frobenius norm

Question

If I have an SVD, $A = X \Sigma Y$, where $\Sigma \in \mathbb{R}^{m \times n}$ has the singular values on the diagonal and $X \in \mathbb{R}^{m \times m}$ and $Y \in \mathbb{R}^{n \times n}$ are orthogonal matrices, then does the squared frobenius norm $||A||^2$ give the summation of the squared singular values of $A$?

Answer 1

Yes. This is a consequence of the invariance of the Frobenius norm under orthogonal transformations.

If $Q$ is an orthogonal matrix, then $\| QA \|_{F}= \| A \|_{F}$. Similarly, $\| AQ \|_{F}=\| A \|_{F}$.

Since $U$ and $V$ in the SVD are orthogonal, $\| A \|_{F}=\| \Sigma \|_{F}$.