$\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\SO}{SO}$ $\newcommand{\eP}{\mathscr{P}}$ $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\tr}{tr}$ Let $m>1$ be an integer and denote by $\eS\_m$ the vector space of symmetric $m\times m$ real matrices. The group $\SO(m)$ acts by conjugation on $\eS\_m$. The space of $\SO(m)$-invariant quadratic polynomials on $\eS\_m$ is spanned by the two polynomials


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


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.



Invariants of symmetric matrices
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 $\mathrm{tr}(A)(A\eta,\eta)$. 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 $\mathrm{tr}(A)$ 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?


