# When does $\Vert AB \Vert = \Vert A \Vert \Vert B \Vert$?

## Motivation

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.

## Question

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 $$\Vert \cdot \Vert = \Vert \cdot \Vert_2$$.

• What is the 2-norm? The answer will change depending on the norm. – Will M. Nov 15 '18 at 23:23
• 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. – ArtificiallyIntelligence Nov 15 '18 at 23:24
• It looks to me you have your definitions missmatched. Check again, you are around a solution, it looks like. – Will M. Nov 15 '18 at 23:26
• 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 $\|BC\| \leqslant \|B\|\|C\|$. – 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 $$\Vert A B\Vert = \sup \Vert AB x\Vert/\Vert x \Vert = \Vert A \Vert \Vert B \Vert$$

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