Simple question about the operator norm

Question

Let $\| \cdot \|$ denote the operator norm of a matrix. It is clear that for all matrices $A, B$ and all vectors $v$, we have $\|ABv\|_2 \leq \|A\|\|Bv\|_2$. Is it also true that $\|ABv\|_2 \leq \|B\|\|Av\|_2$? This is obvious when $A$ and $B$ commute, but I feel it should hold true in general.

Answer 1

No, this is not always true. Consider $$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, $$ and $$ B = \begin{pmatrix} \cos(\pi/2)& -\sin(\pi/2) \\ \sin(\pi/2) & \cos(\pi/2) \end{pmatrix} = \begin{pmatrix} 0& -1 \\ 1 & 0 \end{pmatrix} $$ and the first standard basis vector $v=e_1$: $$ v=e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}. $$ Then we have $\|ABv\|_2 = \|Ae_2\| = \sqrt{2}$, but $\|B\|\|Av\|_2 = 1$. Thus $\|ABv\|_2 > \|B\|\|Av\|_2$.

Discussion. Intuitively, the operator norm of a matrix $\|\cdot\|$ is the "maximum stretch factor" that $B$ imparts to vectors that it sees, so when we do composition in order, we get $\|A(Bv)\|_2$ is bounded by the most that $A$ can stretch a vector times the length of $Bv$. The reason that $\|ABv\|_2$ might be bigger than $\|B\|\|Av\|_2$, intuitively speaking, is that $B$ may not stretch vectors at all, so $\|B\| = 1$, but it may transform a vector $v$ that $A$ does not stretch very much into a position $Bv$ where $A$ stretches it a lot, so $\|ABv\|_2 \gg \|B\|\|Av\|_2=1\cdot\|Av\|_2$!