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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.

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    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?$2011-12-15

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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.