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 $\Vert \cdot \Vert = \Vert \cdot \Vert_2$.
Best Answer
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,
we have $$ \Vert A B\Vert = \sup \Vert AB x\Vert/\Vert x \Vert = \Vert A \Vert \Vert B \Vert $$