Is there a semi-norm that respects matrix similitary?

Question

It is not possible to have a norm on $M_n(\mathbb C)$ that respects similitary for $n > 1$ since, for example, if $A \sim 2A$ and $A \neq 0_n$ then $N(A) = 2N(A)$ contradicts the separating property. For example, $A$ s.t. $A_{1,2} = 1$ and with $0$ elsewhere.

So the question is: Is there a semi-norm (i.e. without the separating property) on $M_n(\mathbb C)$ that respects matrix similitary?

Answer 1

The absolute value of the trace is a seminorm that respects similarity.

Answer 2

Every such semi-norm is a nonnegative constant multiple of the absolute value of the matrix trace.

Suppose $\|\cdot\|$ is a semi-norm that respects similarity. Then $\|N\|=0$ for every nilpotent matrix $N$, because $\|2N\|=\|N\|$ by similarity of $2N$ and $N$ but $\|2N\|=2\|N\|$ by absolute homogeneity of semi-norm.

So, if $A=D+L+U$ where $D,L,U$ are respectively the diagonal, strictly lower triangular and strictly upper triangular parts of $A$, then $\|A\|\le\|D\|+\|L\|+\|U\|=\|D\|$ because $\|L\|=\|U\|=0$. Similarly, as $D=A+(-L)+(-U)$, by triangle inequality we also have $\|D\|\le\|A\|$. Thus $\|A\|=\|D\|$.

The diagonal matrix $D$ can be further decomposed into the sum of $\frac{\operatorname{tr}(D)}nI_n$ and a traceless matrix. However, every traceless matrix is similar to a matrix with a zero diagonal. So, by a similar argument to the one in the previous paragraph, we see that $\|D\|=\left\|\frac{\operatorname{tr}(D)}nI_n\right\|$, meaning that $\|A\|=|\operatorname{tr}(A)|\,\|\frac1nI_n\|$.