What are the semi-norms on $M_n(\mathbb C)$ such that $\| AB\| = \| BA\|$, for all $A$ and $B$? What are the semi-norms on $M_n(\mathbb C)$ such that $\| AB\| = \| BA\|$, for all $A$ and $B$ in $M_n(\mathbb C)$, $n ≥ 2$?
I came across this exercise (an oral exercise), and I have thought about this : Is there a semi-norm that respects matrix similitary?
Do you have another idea to solve this problem? It does not seem really natural to directly think of similitary and does not allow to conclude apparently: are these the only such semi-norms?
 A: Firstly, $\|E_{ij}\|=0$ whenever $i\ne j$ because
$$
\pmatrix{1&0\\ 0&0}\pmatrix{0&1\\ 0&0}=\pmatrix{0&1\\ 0&0}
\text{ and } \pmatrix{0&1\\ 0&0}\pmatrix{1&0\\ 0&0}=0.
$$
It follows that $\|F\|=0$ for any matrix $F$ with a zero diagonal.
Secondly, $\|E_{ii}-E_{i+1,\,i+1}\|=0$ for $i=1,2,\ldots,n-1$ because
\begin{align}
\left\|\pmatrix{1&0\\ 0&-1}\right\|
&=\left\|\frac12\pmatrix{1&1\\ -1&1}\pmatrix{1&1\\ 1&-1}\right\|\\
&=\left\|\frac12\pmatrix{1&1\\ 1&-1}\pmatrix{1&1\\ -1&1}\right\|\\
&=\left\|\pmatrix{0&1\\ 1&0}\right\|\\
&=0\text{ (because the matrix has a zero diagonal)}.
\end{align}
It follows that $\|T\|=0$ whenever $T$ is a traceless diagonal matrix, because $T$ is a linear combination of $E_{ii}-E_{i+1,\,i+1},\ i\in\{1,2,\ldots,n-1\}$. (We have actually used similarity in the above, namely, every traceless matrix is similar to a zero-diagonal matrix, but such use at least is not explicit.)
Now, for any matrix $A$, let $c=\frac{\operatorname{tr}(A)}{n}$. Then $A$ can be split into a sum of the form $cI+T+F$, where $F$ is the off-diagonal part of $A$ and $T$ is a traceless diagonal matrix. Hence
$$
\|A\|=\|cI+T+F\|\le\|cI\|+\|T\|+\|F\|=\|cI\|
$$
and
$$
\|cI\|=\|A-T-F\|\le\|A\|+\|T\|+\|F\|=\|A\|,
$$
meaning that $\|A\|=\|cI\|=\left|\frac{\operatorname{tr}(A)}{n}\right|\|I\|$. In other words, $\|\cdot\|$ must be a scalar multiple of the absolute matrix trace.
