# 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. Commented Nov 15, 2018 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. Commented Nov 15, 2018 at 23:24
• It looks to me you have your definitions missmatched. Check again, you are around a solution, it looks like. Commented Nov 15, 2018 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 $\|BC\| \leqslant \|B\|\|C\|$. Commented Nov 16, 2018 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 $\|A\| = \left( \sum\limits_{p,q} A(p,q)^2 \right)^{\frac{1}{2}}.$ Commented Nov 15, 2018 at 23:22
• Oh, you are OP. Commented Nov 15, 2018 at 23:22