If $a$ and $b$ are vector, then thinking simply vector 2 norm, $\Vert a \cdot b\Vert = \Vert b\Vert \Vert a\Vert \cos(a,b) $, we know the difference is simply a ratio between the angle of $a$ and $b$.

More generally, in a Hilbert space, Cauchy inequality holds so $$|\langle a,b\rangle|^2 \le \langle a,a\rangle\langle b,b\rangle$$ and we know the only when a, b are parallel, the equality is achieved.


Given two square matrix $A$ and $B$,
when does this happen? $$\Vert AB \Vert = \Vert A \Vert \Vert B \Vert$$

Let's simply assume matrix 2-norm, so $\Vert \cdot \Vert = \Vert \cdot \Vert_2$.

  • $\begingroup$ What is the 2-norm? The answer will change depending on the norm. $\endgroup$ – Will M. Nov 15 '18 at 23:23
  • $\begingroup$ 2-norm is just matrix 2-norm or you can say it is an induced norm from the l2 normed vector space. It is the norm from operator sense, treating matrix as an operator. $\endgroup$ – ArtificiallyIntelligence Nov 15 '18 at 23:24
  • $\begingroup$ It looks to me you have your definitions missmatched. Check again, you are around a solution, it looks like. $\endgroup$ – Will M. Nov 15 '18 at 23:26
  • $\begingroup$ Thank you for posting this question! It saved me from a bad mistake in an answer, where I had written (unnecessarily) $\|BC\| = \|B\|\|C\|$, instead of $\|BC\| \leqslant \|B\|\|C\|$. $\endgroup$ – Calum Gilhooley Nov 16 '18 at 1:00

So given $\Vert x \Vert =1$, $$\Vert AB x \Vert \le \Vert A \Vert \Vert Bx \Vert,$$ the equality holds when $Bx$ hit on the direction of first right singular vector of $A$.

Then $$\Vert Bx \Vert \le \Vert B \Vert \Vert x \Vert$$ the equality holds when $x$ hit the first right singular vector of $B$. However, now the $Bx$ aligned with first left singular vector of $B$ and it must match the first right singular vector of $A$.

Note that $$\Vert A B x\Vert \le \Vert A \Vert \Vert B\Vert \Vert x \Vert$$

When both equality conditions are holds, that is to say,

the largest left singular vector of $B$ is parallel to the largest right singular vector of $A$

we have $$ \Vert A B\Vert = \sup \Vert AB x\Vert/\Vert x \Vert = \Vert A \Vert \Vert B \Vert $$

  • $\begingroup$ OP is using 2-norm. I think that signifies $\|A\| = \left( \sum\limits_{p,q} A(p,q)^2 \right)^{\frac{1}{2}}.$ $\endgroup$ – Will M. Nov 15 '18 at 23:22
  • 1
    $\begingroup$ Oh, you are OP. $\endgroup$ – Will M. Nov 15 '18 at 23:22

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