# Are matrix $p$-norms unitary invariant?

Consider a matrix $X \in \mathbb{R}^{N \times N}$. Let $\| X \|_p$ be its $p$-norm

$$\| X\|_p = \left( \sum_{ij} |X_{ij}|^p \right)^{\frac 1p}$$

Is $\|X\|_p$ unitary invariant? That is, given any unitary matrix $U$, $V$, is $\|X\|_p$ = $\|UXV\|_p$?

My strategy is to formulate the $p$-norm into some kind of singular value or trace, but I failed to do this. If it is not unitary invariant, how can I prove it?

• The first thing I would do here is to check, whether random unitary matrices have the same $p$-norm as the identity matrix. Have you done that? Note: in the case of real matrices I would use "orthogonal" in place of "unitary". The latter term IMHO screams "complex entries possible". Aug 16, 2013 at 5:06
• @JyrkiLahtonen Sorry I forget to compute it on an arbitrary case first. Its not unitary invariant. I will close this question.
– Rein
Aug 16, 2013 at 5:22
• .. or you can answer it with by giving a single counterexample. Nothing to worry. I've (and most of us) done worse booboo. Aug 16, 2013 at 5:24

I suppose you already know about $p=2$ which is the frobenius norm of the matrix and is the only norm which is unitarily invariant among all $p$. To prove it is unitarily invariant, try expressing the frobenius norm in terms of trace operator and then use the circularity of trace. To prove other cases are not unitarily invariant, try picking examples for any given $p$. I don't know how to prove it. I am checking on it.
Update: First of all, it is enough to prove that vector $p-$ norms are not unitarily invariant. This follows from the factor that matrix $p-$ norms can be expressed as vector $p-$norms of its columns. The idea is to prove that for any given $p\neq 2$, you can find a vector and a unitary matrix such that unitary invariance doesn't hold. Take $x=[1,0,\dots,0]^T$. Note that $(||x||_p)^p=1$. Thus, we need to find a unitary matrix $U$ such that $(||Ux||_p)^p\neq 1$. Consider a unitary $U$ whose first column is given by $c_1=\left[\frac{1}{\sqrt{N}},\frac{1}{\sqrt{N}},\dots,\frac{1}{\sqrt{N}}\right]$ (eg: DFT Matrix). Now, note that $(||c_1||_p)^p=N^{1-\frac{p}{2}}$(note the case $p=2$). Now observe that $(||Ux||_p)^p=(||c_1||_p)^p\neq 1 = (||x||_p)^p$. Thus for every $p$ except $p=2$, you can construct a counter example.
• Please note that in the comments, the OP has already concluded that the question is not true for any $p$ and is going to close it. It was a comment to that. Upon your comment, I have expanded the answer Aug 16, 2013 at 9:22
• As it was before, it was a comment (and that's what I clicked in review, when it posted "my" comment above). The way it is now, I think it qualifies for an answer and an upvote (especially if you show that it's not unitary invariant for $p \ne 2$). Aug 16, 2013 at 11:23