Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M_n(\mathbb C)$ be the algebra of $n\times n$ complex matrices. The coefficients of the characteristic polynomial $\det(\lambda I-A)=\sum f_i(A)\lambda^i$ are polynomials in the entries of $A\in M_n(\mathbb C)$, and are conjugation invariant. Moreover, every conjugation invariant polynomial function $F:M_n(\mathbb C)\to \mathbb C$ is of the form $P(f_0,f_1,\ldots, f_{n-1})$ for $P\in \mathbb C[x_0,\ldots,x_{n-1}]$.

Here is a proof (hover mouse over to view):

Given a group action on a space, any continuous invariant function must be constant on the closure of each orbit. Because the closure of every conjugacy class contains a diagonal matrix, the entries of which are the eigenvalues of the matrices in the conjugacy class, invariant functions must be polynomials in the eigenvalues. Given any permutation $\sigma\in S_n$, we have that $\operatorname{diag}(a_1,\ldots, a_n)$ is conjugate to $\operatorname{diag}(a_{\sigma(1)},\ldots, a_{\sigma(n)})$, and hence an invariant function must be a symmetric function in the eigenvalues. The coefficient $f_i(A)$ is up to a sign the $(n-i)$th elementary symmetric function in the eigenvalues of $A$, and since the elementary symmetric functions generate the ring of all symmetric functions, the result follows.

Is there a proof that doesn't require reducing the problem to eigenvalues and symmetric functions? Perhaps more important, can the result be extended to non-algebraically closed fields or other base rings where this particular proof fails because we don't have diagonalization? If not, what additional conjugation-invariant polynomial functions are there over $\mathbb R$, $\mathbb Q$, or $\mathbb Z$?

share|cite|improve this question
To clarify: A polynomial function $F:M_n(\mathbb C)\to \mathbb C$ is a polynomial in the entries of its argument? – joriki Nov 2 '11 at 5:36
@Joriki: Yes. I believe this is equivalent to viewing $M_n(\mathbb C)$ and $\mathbb C$ as algebraic varieties (affine spaces) and considering regular functions between them. – Aaron Nov 2 '11 at 5:43
Dear @joriki: I see that Aaron has already responded, but here is the (more naive) comment I prepared: Yes. You view $M_n(\mathbb C)$ as a (finite dimensional) vector space. – Pierre-Yves Gaillard Nov 2 '11 at 5:49
Dear Aaron: I answered your question as well as I could. I'm sure there will be better answers, and I'm expecting them with as much impatience as you. - I don't have the slightest idea about the conjugacy classes in $M_n(\mathbb Z)$. That's definitely too complicated form me! (Thanks for your very interesting question, which I upvoted.) – Pierre-Yves Gaillard Nov 2 '11 at 9:54
@Aaron: the proof using eigenvalues extends to non-algebraically closed fields (take the algebraic closure and extend a given polynomial function to the algebraic closure). Is this a sufficiently satisfying answer to your question? – Qiaochu Yuan Apr 30 '12 at 21:20

Let $K$ be an infinite field and $f:M_n(K)\to K$ a polynomial map which is constant on the conjugacy classes.

Claim. $f$ is a polynomial in the coefficients of the characteristic polynomial.

We can (and will) assume that $K$ is algebraically closed.

Observation. If a polynomial map $g:M_n(K)\to K$ vanishes on the diagonalizable matrices, then $g=0$.

Proof of the observation. We have $dg=0$, where $d$ is the discriminant of the characteristic polynomial. As $K$ is infinite, this implies $g=0$. QED

Proof of the claim.

By the observation, $f$ is determined by its restriction to the diagonal matrices.

This restriction is invariant by permutation of the diagonal entries.

Thus, this restriction is a polynomial in the elementary symmetric polynomials in the diagonal entries.

But these elementary symmetric polynomials are the coefficients of the characteristic polynomial.

Using again the observation, we see that $f$ itself is a polynomial in the coefficients of the characteristic polynomial. QED

share|cite|improve this answer
This seems very similar to the proof in the question -- it uses algebraic closure, topology and diagonalizability. I understood the question to be asking for a proof that doesn't use those and might apply to $\mathbb R$, $\mathbb Q$ and even $\mathbb Z$. – joriki Nov 2 '11 at 8:18
Dear @joriki: Thanks for your comment. I'm planning to edit the answer. I'll try to prove the statement for an infinite field, without using any word having a topological flavor. Traditionally, polynomial maps are not considered over finite fields. I'm definitely unable to prove the statement over $\mathbb Z$. (Do you think it holds over $\mathbb Z$?) – Pierre-Yves Gaillard Nov 2 '11 at 8:45
@Pierre-YvesGaillard: Thanks for the proof. It is very similar in spirit to my proof, except that you prove the observation in a more algebraic way than I do. However, could you expand your proof of the observation a little? I'm not sure how dg initially enters the picture. Additionally, while I don't know anything about conjugacy classes over Z, I found some notes which talk about how to find the number of conjugacy classes with a given (irreducible) characteristic polynomial, and may have more insight to offer. – Aaron Nov 2 '11 at 15:10
Dear @Aaron: You're welcome. Thanks for your comment. I agree with what you say. I'll take a look at Keith Conrad notes. (By the way, he's an MSE user. See, below this question, the comment to starting with @KCd: Dear Keith. I'm sure he'd have interesting things to say on your questions.) For the observation, if $d(A)\neq0$, then $A$, having $n$ distinct eigenvalues, is diagonalizable, and we have $g(A)=0$. – Pierre-Yves Gaillard Nov 2 '11 at 15:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.