Example of matrices that do not satisfy the submultiplicative property In rare cases there are some matrices that do not satisfy the submultiplicative property:
$$\|AB\| \leq \|A\| \|B\|$$
Is there an example of such a matrix $A$ and $B$ such that
$$\|AB\| > \|A\| \|B\| \quad ?$$
 A: We can define a norm on matrices $M=(m_{ij})$ over $\mathbb{R}$ by $$||M||=\max_{i,j}(|m_{ij}|).$$  We can readily verify this is a matrix norm by checking the definitions hold.
So, if we let
$$
A=\begin{pmatrix}
2 & 2 \\
2 & 2 \\
\end{pmatrix}
$$
then
$$
A^2=\begin{pmatrix}
8 & 8 \\
8 & 8 \\
\end{pmatrix}
$$
and so $$||A \cdot A||=8>4=||A|| \cdot ||A||.$$
A: I guess that many generalizations of vector norms to matrices are good candidates for matrix norms which do not satisfy the submultiplicative property. In particular, norms of the form
$$
|||A|||_p = \left(\sum_{i,j}|a_{ij}|^p\right)^{1/p}, \quad 1 \leq p\leq\infty
$$
(they are norms obviously since they are defined as the vector $p$-norms of the matrix reshaped to a vector).
The case $p=\infty$ (in the limit sense of course) was pointed out by Rebecca and it is used as the most common example of the matrix norm which do not satisfy the property. 
I am not sure where this was considered before but my feeling is, that the submultiplicative property does not hold if $p>2$ while, on the other hand and more interestingly, for $p\leq 2$ the property holds (with the most obvious case with $p=2$ being the Frobenius norm). You can illustrate the first part similarly as for the max-norm by taking 
$$ A = B = \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix} $$
with $|||A|||_p = |||B|||_p = 4^{1/p}$ and $|||AB|||_p = 2\cdot 4^{1/p}$. Obviously for $p>2$ you have $|||AB|||_p > |||A|||_p|||B|||_p$.
EDIT: I guess that instead of "In rare cases there are some matrices that do not satisfy the submultiplicative property", the actual question would be "Are there some (rare) matrix norms which do not satisfy the submultiplicative property?" I would certainly be interested in more examples of them than those mentioned above :-)
A: If you are using the operator norm on matrices, then it is not possible. The operator norm is submultiplicative and makes the set of matrices into a Banach algebra.
What I mean to say is that $||AB|| \leq ||A||\cdot||B||$ for all matrices $A$ and $B$.
You can look up operator norm on Wikipedia.
A: Assuming $\|A\|$ is the operator norm of $A$:$\|Ax\|\le \|A\|\cdot \|x\|$, and $\|A\|$ is the least such constant, then the answer is no, since it's a theorem (easily proved from the definition I just gave) that $\|AB\|\le \|A\|\cdot \|B\|$ always holds. Moreover, I'm not sure what you mean by rare cases, but it certainly happens very often that this inequality is not an equality. It is the case though that this inequality can't be improved (without any extra assumptions on the matrices). 
