one question about $\|V^HAV\|=\|A\|$

Question

I have one question about matrix norm:

If $V$ is an nonsquare orthogonal matrix, $V\in C^{n\times m}$ and A is a square matrix, $A\in C^{n\times n}$.

Is it correct that $\|V^HAV\|=\|A\|$.

If correct, please show the reason or proof.

Thank you

Answer 1

This is more or less true for square unitary matrices, but if $V$ is not square, then this isn't true.

You haven't specified what matrix norm you want to use. As it happens, both the Frobenius norm ($\| A \|_{F}$), and the spectral or 2-norm ($\| A \|_{2}$) are invariant under unitary transformations. For these norms, if $V$ is unitary, then $\| A \|=\| V^{H}A \|$ and $\| A \|= \| AV \|$.

To see this for the 2-norm, recall that

$\| A \|=\sqrt{\lambda_{\max}(A^{H}A)}$

Thus

$\| V^{H}A \|_{2}=\sqrt{\lambda_{\max}(A^{H}VV^{H}A)}=\sqrt{\lambda_{\max}(A^{H}A)}=\| A \|_{2} $.

For the second part, recall that eigenvalues are invariant under unitary similarity transformations. Thus

$\| AV \|_{2}=\sqrt{\lambda_{\max}(V^{H}(A^{H}A) V)}=\sqrt{\lambda_{\max}(A^{H}A)}=\| A \|_{2}$.

Answer 2

This is not true. Take $$V=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ and $A=I$ (identity matrix). By your terms, $V$ is a "non-square" orthogonal matrix (there is only one orthogonal vector though). But if it is a square, orthonormal matrix, this will hold true for both frobenius norm and induced 2-norm. For instance $$||V^H AV||_2 = \max_{||x||_2=1}||V^HAVx||_2 = \max_{||y||_2=1}||V^HAy||_2=\max_{||y||_2=1}||Ay||_2$$ This holds true because, $||Vx||_2 = ||x||_2$ for all orthonormal matrices $V$ (try proving yourself). A similar argument hold for frobenius norms and you can prove it using the trick $||A||_F^2=trace\{A^HA\}$