Fix some integer $n$, and consider the linear space $M(n,\mathbb F)$ of square $n\times n$ matrices in some field $\mathbb F$. Let $f:M(n,\mathbb F)\to\mathbb F$ be a functional that is invariant under change of basis, that is, such that $f(PAP^{-1})=f(A)$ for any $A,P\in M(n,\mathbb F)$ with $P$ invertible.
Standard examples are $f(A)=\det(A)$ and $f(A)=\operatorname{tr}(A)$. More generally, any function defined via the eigenvalues of $A$ is another example of this.
Are these the only possible such examples?
In other words, can we characterise the set of possible functionals $M(n,\mathbb F)\to\mathbb F$ that are invariant under change of basis as being all and only those functions that can be defined from the eigenvalues of the matrix? I'm mostly interested to the cases $\mathbb F=\mathbb R,\mathbb C$, but I'm leaving this question general because I don't know if this assumption is relevant to the discussion.
Best Answer
Here is an observation that you might find interesting.
Proof of claim: I will prove this for $\Bbb F = \Bbb C$, but the proof for $\Bbb F = \Bbb R$ is similar. Define $g:\Bbb C^n \to \Bbb C$ by $$ g(\lambda_1,\dots,\lambda_n) = f (\operatorname{diag}(\lambda_1,\dots,\lambda_n)), $$ where $\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ is the diagonal matrix with diagonal entries $\lambda_1,\dots,\lambda_n$. It is clear that if $A$ is diagonlizable, then $f(A) = g(\lambda_1(A),\dots,\lambda_n(A))$. Suppose that $A$ is not diagonalizable. Let $J = PAP^{-1}$ be the Jordan form of $A$. Define $$ D_k = \operatorname{diag}(1,k,\dots,k^n). $$ Note that because $J$ is upper-triangular, $\lim_{k \to \infty}D_kJD_k^{-1} = \Lambda$, where $\Lambda$ denotes the diagonal matrix whose diagonal is equal to that of $J$. Conclude that $$ g(\lambda_1(A),\dots,\lambda_n(A)) = f(\Lambda) = f\left(\lim_{k \to \infty} D_kJD_k^{-1}\right) \\ = \lim_{k \to \infty} f(D_kJD_k^{-1}) = \lim_{k \to \infty} f(J) = f(J) = f(A). $$