Simple question about the operator norm 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.
 A: 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 $\|ABv\|_2 \gg \|B\|\|Av\|_2=1\cdot\|Av\|_2$!
