# Invariants of a matrix

I'm teaching a course in physics, and I need a simple and intuitive proof that a matrix ($3\times3$, but it doesn't matter) has exactly 1 invariant which is linear in its entries, 2 that are quadratic, etc. When I say "invariance" I mean under orthonormal transformation of the axes $A\to Q A Q^{-1}$ for orthonormal $Q$.

For example, that every quadratic invariant scalar is a combination of $\operatorname{tr}(A)^2$ and $\operatorname{tr}(A^2)$.

The proof for the linear case is trivial (and intuitive) but I can't find a generalization for the quadratic case.

• Do you have a (non-intuitive) proof that the resutl is true in the first place? Writing $p_k$ for $\mathop{tr}(A^k)$, which are invariant even under conjugation by any invertible matrix, I find in dimension $4$ the $5$ linearly independent invariants $p_4$, $p^3p_1$, $p_2^2$, $p_2p_1^2$, $p_1^4$, which contradicts your claim. In general one gets the number of partitions of $n$ this way, which grows fast. Dec 15, 2011 at 13:54
• I don't see your point. $p_4$ is fourth order in the entries of $A$. Dec 15, 2011 at 14:15
• A proof for the linear case: say that $f(A)$ is a scalar invariant that is linear in the entries of $A$. That means $$f=\sum_{ij} C_{ij}A_{ij}$$ where $C$ is some matrix. $C$ should be unchanged when applying an infinitesimal rotation to $A$. This means (here there're two lines of algebra) that $C$ commutes with the generators of $SO(n)$. The only $C$ that does that is the identity. QED Dec 15, 2011 at 14:15
• The point is that you generalized this by "etc.", and the counterexample shows that it doesn't hold for quartic invariants. It may still well hold for quadratic invariants. Dec 15, 2011 at 14:18
• @yohBS: It is known (and easy to see) that the polynomial invariants under conjugation by $GL(n)$ are precisely the polynomials in the coefficients of the characteristic polynomial, which in characteristic $0$ is the same as polynomials in the $p_k$. So your question would be: "are the polynomial invariants for conjugation by $O(n)$ automatically invariants for conjugation by $GL(n)$?" If that is what you mean, it would be clearer to ask the question that way (and also it seems unlikely to this would be true). Dec 15, 2011 at 16:19

This is false already for quadratic invariants, since $\operatorname{tr}A^\top A$, the square of the Frobenius norm, is invariant under orthogonal transformations,
$$\operatorname{tr}(QAQ^\top)^\top QAQ^\top=\operatorname{tr}QA^\top Q^\top QAQ^\top=\operatorname{tr}A^\top A\;,$$
and isn't generated by $(\operatorname{tr}A)^2$ and $\operatorname{tr}(A^2)$, since it contains terms $A_{ij}^2$ with $i\ne j$ that don't occur in either of the two.
A full theory of polynomial matrix invariants under $SO(n)$ is developed in this paper. It turns out that the traces of all products of $A$ and $A^\top$ generate all polynomial invariants under $O(n)$, but there are additional more complicated invariants under $SO(n)$.