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I came across this inequality

$$\|ABC\|_F \leq \|A\| \|B\|_F \|C\|$$

for all matrices $A$, $B$ and $C$, where $\|\cdot\|$ is the operator norm (max singular value)

I do not know how to prove this, nor do I have it accessible. Is there a proof available online?

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1 Answer 1

up vote 2 down vote accepted

The key fact is that $\|A^*A\|\,I-A^*A$ is positive-semidefinite, and then so is $B^*(\|A^*A\|-A^*A)B=\|A^*A\|B^*B-B^*A^*AB$.

Then, as the eigenvalues of a positive-semidefinite matrix are all non-negative, we have $$ \|AB\|_F^2=\mbox{Tr}((AB)^*AB)=\mbox{Tr}(B^*A^*AB)\leq\mbox{Tr}(\|A^*A\|B^*B)=\|A^*A\|\,\mbox{Tr}(B^*B)\\ =\|A^*A\|\,\|B\|_F^2=\|A\|^2\,\|B\|_F^2. $$ Now, $$ \|ABC\|_F=\|A(BC)\|_F\leq\|A\|\,\|BC\|_F=\|A\|\,\|C^*B^*\|_F\leq\|A\|\,\|C^*\|\,\|B^*\|_F =\|A\|\,\|B\|_F\,\|C\|. $$

$\ \\$

Edit: here's a short proof of the fact in the first line. As $A^*A$ is positive-semidefinite, it is diagonalizable, i.e. $A^*A=VDV^*$ for a unitary $V$ and diagonal $D$. As $\|A^*A\|=\|D\|$, it is clear that $\|A^*A\|\,I-D$ is positive-semidefinite (it is a diagonal matrix with non-negative diagonal entries). Then $$ \|A^*A\|\,I-A^*A=\|A^*A\|\,VV^*-VDV^*=V(\|A^*A\|\,I-D)V^* $$ is positive-semidefinite.

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thanks ! I think I understand. One question though: how do you get the inequality in the first line of the proof? i.e: Tr(B*A*AB) \leq Tr(||A*A||B*B) ? –  NSR Aug 30 '12 at 20:15
    
@NSR: the trace is positive-preserving, so if you apply the trace to the positive-semidefinite matrix $\|A^*A\|B^*B-B^*A^*AB$, you get a non-negative number. Coupled with the fact that the trace is linear, this gives you the inequality. –  Martin Argerami Aug 30 '12 at 20:27
    
oh ok... thanks again ! –  NSR Aug 30 '12 at 20:30

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