Invariants of symmetric matrices


$$A \mapsto \tr A^2,\;\;A\mapsto (\tr A)^2.$$

Let $\SO(m-1)$ be the subgroup of $\SO(m)$ consisting of orthogonal transformations of $\bR^m$ that fix a unit vector $\eta$.

What is the space of $\SO(m-1)$-invariant on $\eS_m$. More precisely, can one explicitly write a basis of this space?

Clearly $\tr A^2$ and $(\tr A)^2$ are such polynomials, and so are

$$A\mapsto (A^2\eta,\eta),\;\;A\mapsto (A\eta,\eta)^2,$$

where $(-,-)$ denotes the natural inner product on $\bR^m$. Are there any more $\SO(m-1)$ invariant quadratic polynomials? (I am inclined to believe that the above is the complete list.) Update After Robert Bryant's answer I lost my initial inclination.

-

This is a trick question, right? It's not true when $m=2$ because, then $\mathrm{SO}(m{-}1)=\mathrm{SO}(1)$ is trivial, so that all polynomials on $2$-by-$2$ symmetric matrices are invariants, and the four quadratics you mention clearly don't span the the six-dimensional space of all quadratics.

Moreover, for $m>2$, you are missing $\tr(A)(A\eta,\eta)$. When $m>2$, these five do span the space of $\mathrm{SO}(m{-}1)$-invariant quadratic polynomials, as you can see by realizing that, when $m>2$, the space $\mathcal{S}_m$ of symmetric $m$-by-$m$ matrices splits under $\mathrm{SO}(m{-}1)$ into a direct sum $$\mathcal{S}_m = \mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}^{m-1}\oplus \mathcal{S}^0_{m-1}$$ of $\mathrm{SO}(m{-}1)$-irreducible modules, where the only equivalences are between the first two (trivial) summands, and where $\mathcal{S}^0_{m-1}$ means the trace-zero symmetric $(m{-}1)$-by-$(m{-}1)$ matrices. (These two projections to $\mathbb{R}$ are given by the invariant linear forms $\tr(A)$ and $(A\eta,\eta)$.) [This splitting is valid for $m=2$, of course, but then the third summand is another copy of $\mathbb{R}$ and the fourth summand has dimension $0$.]

However, you switched from quadratic polynomials at the beginning to all polynomials. Was that intentional, or did you just want quadratics?

-
@ Robert It was not a trick question. I wrote it early in the morning when my brain hadn't yet absorbed all the caffeine. I meant quadratic polynomials, and I now see that missed one. Thank you for the answer. –  Liviu Nicolaescu Aug 15 '12 at 14:32