norm of non-square matrix I have the question about the non-square matrix.
If $A$ is $m \times n$-matrix, the $m$ is not equal $n$. $x$ is a vector which the dimension is $n \times 1$.
Is it correct that 
   $$\left\Vert Ax \right\Vert \leq \left\Vert A\right\Vert \left\Vert x\right\Vert ?$$
If $A$ is $m \times n$-matrix and $B$ is $n \times m$-matrix, is it correct that
   $$\left\Vert AB \right\Vert \leq \left\Vert A\right\Vert \left \Vert B \right\Vert?$$
Note: I'm mainly ask about norms like $\|\cdot\|_1$, $\|\cdot\|_2$, $\|\cdot\|_\infty$
 A: For a vector norm $\|\cdot\|_\star$ (e.g. $\star=1,2,\infty$) one define an induced matrix norm by
$$\|A\|_\star=\max_{\|x\|_\star\ne 0}\frac{\|A x\|_\star}{\|x\|_\star}$$
for any finit dimentional matrix $A$.
it can be proved from that the following is equivalent definition
$$\|A\|_\star=\max_{\|x\|_\star=1}{\|A x\|_\star}$$
Let $x^*$ be a vector such that $\|x^*\|_\star=1$ and
$$\max_{\|x\|_\star=1}{\|A x\|_\star}=\|A x^*\|_\star,$$ 
then for any other $\|x\|_\star=1$ we have 
$$\|A x\|_\star\leq\|A x^*\|_\star=\|A\|_\star=\|A\|_\star\|x\|_\star$$
Now $\|ABx\|_\star\leq \|A\|_\star\|Bx\|_\star\leq \|A\|_\star\|B\|_\star\|x\|_\star$ , therefore $$\|AB\|=\max_{\|x\|_\star=1}{\|A B x\|_\star}\leq
\max_{\|x\|_\star=1}{\|A\|\|B\|\|x\|_\star}=
\|A\|_\star\|B\|_\star$$
Note that nothing in the above required that the matrices needs to be square, so it works for any $n\times m$ matrix.
By the way, for the Frobeneous Norm which isn't an induced norm these properties still holds. I guess for a general non induced norm this may and may not work.
A: Shortly, yes, the inequality holds. In details...
Generally, let $A: V \to W$ be linear map between linear spaces $V$ with norm $\Vert \cdot \Vert_V$ and $W$ with norm $\Vert \cdot \Vert_W$. Then you can define operator norm (though the name may be different in different books) as
$$
\Vert A \Vert_{W,V} = \sup_{x  \neq 0 , \in V} \frac{ \Vert Ax \Vert_W }{ \Vert x \Vert_V }.
$$
From here you can easily prove assuming  all norms to be finite that for all $x\in V$
$$
\Vert Ax \Vert_W \leq \Vert A \Vert_{W,V} \cdot \Vert x\Vert_V
$$
and then prove that for $B:W \to V$
$$
\Vert AB \Vert_W \leq \Vert A \Vert_{W,V} \cdot \Vert B \Vert_{V,W}
$$
which is what you need.
