6

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.

Zev Chonoles
  • 124,702
  • 19
  • 297
  • 503
yohBS
  • 2,808
  • 17
  • 24
  • 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. – Marc van Leeuwen Dec 15 '11 at 13:54
  • I don't see your point. $p_4$ is fourth order in the entries of $A$. – yohBS Dec 15 '11 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 – yohBS Dec 15 '11 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. – joriki Dec 15 '11 at 14:18
  • Agreed. I meant that all invariants may be expressed by linear combinations of $p_{i_1}^{k_1}\dots p_{i_m}^{k_m}$. But clearly there may be more than $n$ of these in the $n$th order. – yohBS Dec 15 '11 at 14:19
  • You used the assumption that if tr$CA=$tr$C'A$ then $C=C'$. That is not generally true, is it? – yohBS Dec 15 '11 at 14:28
  • I used the assumption that $C=C'$ if $\operatorname{tr}CA=\operatorname{tr}C'A$ for all $A$. That's true because we can vary the $A_{ij}$ independently so their coefficients $C_{ij}$ must be equal independently. – joriki Dec 15 '11 at 14:32
  • Sorry -- although the above comment is correct, the proof also relied on the fact that $C$ has to commute with orthonormal matrices, so it wasn't actually more direct; I've deleted it. – joriki Dec 15 '11 at 14:42
  • Cross-posted on P.SE http://physics.stackexchange.com/q/18321/2451 – Qmechanic Dec 15 '11 at 16:17
  • 1
    @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). – Marc van Leeuwen Dec 15 '11 at 16:19
  • I see one direction - clearly the coefficients of the characteristic polynomial are invariant under $GL(n)$ (and hence under $O(n)$). But how is it easy to see that these are the only ones, or that these are the polynomials in the $p_k?$ – yohBS Dec 15 '11 at 22:43

1 Answers1

8

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 $SO(n)$.

You may also be interested in this question at MO.

joriki
  • 215,929
  • 14
  • 263
  • 474