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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.

glS
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    See https://math.stackexchange.com/questions/181237/the-stucture-of-invariant-polynomials-on-matrix for an answer under the assumption that $f$ is polynomial. If you don't assume that $f$ is polynomial then there are other examples. For example, $f(A) = \textrm{rank}(A^2)$ is invariant but is not a "function of the eigenvalues". – levap Jul 13 '21 at 22:17
  • If you are talking about arbitrary functions then the question is kind of silly--as pointed out by the answer, all you are asking for is an arbitrary function on the set of equivalence classes of matrices with respect to conjugation (so your question about eigenvalues is just asking whether any two matrices with the same eigenvalues are conjugate, which I imagine you already know is false). A more interesting question would be if you restricted to something like polynomial functions. – Eric Wofsey Jul 13 '21 at 23:41
  • Take two non-conjugate matrices $A$ and $B$ which have the same eigenvalues. Define $f$ which is $1$ on the equivalence class of $A$ and $0$ on the equivalence class of $B$ (and whatever you want on the other equivalence classes). – Eric Wofsey Jul 13 '21 at 23:46
  • @EricWofsey sorry, didn't see the edit to the first comment. That's a very good point about dissimilar matrices with same eigenvalues. Makes me wonder whether rather then looking at all eigenvalues, one should consider the statement for something like eigenvalues counted only with their geometric multiplicity, – glS Jul 14 '21 at 00:13
  • @EricWofsey so, upon reflection, given that conjugacy classes are characterised by the eigenvalues with their geometric multiplicity (https://math.stackexchange.com/q/8339/173147), it is trivial to say that any function of conjugacy classes factorises through these, correct? Because such sets of eigenvalues are in bijection with the conjugacy classes. – glS Jul 14 '21 at 10:03
  • @glS: The geometric multiplicities (and algebraic multiplicities) are not enough; you have to actually know the size of each Jordan block. For instance, you could have an eigenvalue with Jordan blocks of size 1 and 3 or size 2 and 2, and in both cases the geometric multiplicity is 2 and algebraic multiplicity is 4. – Eric Wofsey Jul 14 '21 at 16:00

2 Answers2

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Here is an observation that you might find interesting.

Claim: Let $\Bbb F \in \{\Bbb R, \Bbb C\}$. If $f:M(n,\Bbb F) \to \Bbb F$ is invariant under change of basis and continuous, then there exists a (symmetric) function $g$ such that $f(A) = g(\lambda_1(A),\dots,\lambda_n(A))$.

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 $$ g(\lambda_1(A),\dots,\lambda_n(A)) = f(\Lambda) = f\left(\lim_{k \to \infty} D_kJD_k^{-1}\right) \\ = \lim_{k \to \infty} f(D_kJD_k^{-1}) = \lim_{k \to \infty} f(J) = f(J) = f(A). $$

Ben Grossmann
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  • well now that's fascinating. Am I reading this correctly that this is saying/using that, whereas matrices with same eigenvalues but different Jordan structures are not similar, the closure of the diagonalisable matrices with eigenvalues ${\lambda_i}$ equals the set of all matrices with said eigenvalues? – glS Jul 14 '21 at 20:01
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    @glS That's not quite the correct statement. What is true is that for any particular Jordan form, the closure of the set of matrices with that Jordan form will include the set of diagonalizable matrices with those same eigenvalues. – Ben Grossmann Jul 14 '21 at 20:09
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A function $f:\mathcal M_n(\mathbb F) \to \mathbb F$ is invariant under $A\to P^{-1}AP$ if it is constant on the equivalence classes of similar matrices. A complete set of invariant for matrix similarity is given by the so-called invariants de similitude (I don't know how they are called in english). So $f$ is invariant if, and only if, it factorizes as a function of those invariants.

This includes the determinant, the trace, as well as any function of the characteristic polynomial, but also many others.

SolubleFish
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    I think you answer will be more clear if you show an example of such functional that is not any of those pointed by the OP. But you answer was very clarifing! – DiegoMath Jul 13 '21 at 22:03
  • According to Wiki's "Change Language" links, invariants de similitude are the "tensor invariants". – Jacob Manaker Jul 13 '21 at 22:32
  • these "tensor invariants" seem to be always only defined in terms of the eigenvalues. Does the French version of the page say something different? – glS Jul 14 '21 at 09:22