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For something I'm working on, I have a matrix $A$ with other matrices $U$ and $V$ which are unitary ($U^*U = I$ and $V^*V = I$), and I'm trying to show that, for the Frobenius norm, $\|A\| =\|UA\| =\|AV\| = \|UAV\|$. Now, I solve out the first portion, but everything else is giving me trouble (the 3rd and 4th parts of the equation). Do I have to make use of singular decomposition somehow?

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  • $\begingroup$ Recall $\| B\|^2=\mathrm{tr}(B^*B)$. And the trace is commutative. $\endgroup$
    – Julien
    Nov 13, 2013 at 17:18
  • $\begingroup$ So, in general, $tr(B^*B) = tr(BB^*)?$ $\endgroup$
    – Incognito
    Nov 13, 2013 at 17:21
  • $\begingroup$ Even more generally, the trace of $AB$ is equal to the trace of $BA$. $\endgroup$
    – Julien
    Nov 13, 2013 at 17:23
  • $\begingroup$ I initially (as in many years ago!) found the trace definition to be a little magical but rather vague. $\endgroup$
    – copper.hat
    Nov 13, 2013 at 17:27
  • $\begingroup$ @copper.hat A quality of the trace definition is that it makes it clear that the Frobenius norm is inherited from an inner product. And therefore the generalization to the Schatten norms and their natural duality like that of $\ell^p$ and $\ell^q$. $\endgroup$
    – Julien
    Nov 13, 2013 at 18:31

1 Answer 1

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If $U$ are unitary, then $\|U x\| = \|x\|$.

Then we have $\|A\|_F^2 = \sum_k \|A e_k\|_2^2 = \sum_k \|UA e_k\|_2^2 = \|UA\|_F^2$

We also have $\|A\|_F^2 = \|A^*\|_F^2$. The above result gives $\|A\|_F^2 = \|A^*\|_F^2 = \|V^*A^*\|_F^2 = \|AV\|_F^2$.

Combining gives the desired result.

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  • $\begingroup$ Sure, you may find this more natural. And actually, that's the way Hilbert-Schmidt operators are usually introduced so I agree. I just assumed that the OP was given the trace definition. For otherwise, proving $\|AV\|=\|A\|$ is actually proving that $\|A\|$ is well-defined and independent of the choice of an orthonormal basis. So it is either circular or tautological from that perspective. $\endgroup$
    – Julien
    Nov 13, 2013 at 18:43

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