For general non-symmetric square matrices is there a matrix norm that is invariant under similarity transformations? I think that there is no similarity-invariant matrix norm for general matrices. But are there similarity invariant norms for special types of matrices (e.g. for matrices whose eigevalues are different from zero)? Can these norms be derived from an inner product?
 A: IMHO for $N(\cdot)$ to be a unitarily invariant norm it would be necessary that $N$ depends entirely on the eigenvalues of its argument. I don't think this is possible for general matrices. If such a norm existed then a zero matrix and a triangular matrix with zeros on the diagonal would necessarily need to have the same norm since they have the same spectrum. This is however not possible since $N(X)=0$ can happen if and only if $X=0$.
Some very special cases are possible though: e.g., symmetric or Hermitian matrices, which even form a linear subspace of $\mathbb{R}^{n\times n}$ or $\mathbb{C}^{n\times n}$. For such matrices, the usual spectral and Frobenius norms (or any other norm which can be written in terms of the singular values of the matrix) depend entirely on the spectrum (simply because the singular values are just the absolute values of the eigenvalues in this case).
A: I think it can be proved like this.
If I well understood the question is are there norms on $M(n)$ such that $\|A\| = \|C^{-1}AC\|$ for every invertible $C$ and every $A$. If $\|\cdot\|$ is scuh norm, than we have $\|AC\| = \|CA\|$, for every $A$ and every invertible $C$. But invertible matrices are dense in $M(n)$, for example in the operator norm, and since all norms are equivalent on the finite dimensional vector space, they are also dense in the norm $\|\cdot\|$, which implies $\|AC\| = \|CA\|$ for every $A,C\in M(n)$. But we can found $A,C$ such that $AC = 0$ and $CA\neq 0$, which is a contradiction
