Eigenspaces in the symmetric power Given a finite dimensional vector space $V$ and a linear map $f:V\to V$, we can construct a multilinear map, which we will denote by $F$, on the $n^{th}$ symmetric power of $V$, which I tend to think of as a degree $n$ polynomial in a chosen basis of $V$, by applying $f$ to each copy of $V$.
Suppose we understand the eigenspaces of $f$ explicitly. Can we then construct the eigenspaces of $F$ from those of $f$? Something like "every eigenvector of $F$ is a polynomial in eigenvectors of $f$" or the like.
 A: Generally, if $f : V \to V$ and $g : W \to W$ are two linear maps, the tensor product $f \otimes g : V \otimes W \to V \otimes W$ satisfies $(f \otimes g)(v \otimes w) = f(v) \otimes g(w)$. This implies that if $v_i$ is an eigenvector of $f$ with eigenvalue $\lambda_i$ then
$$\begin{align*} f^{\otimes k}(v_{i_1} \otimes \dots \otimes v_{i_k}) 
 &= f(v_{i_1}) \otimes \dots \otimes f(v_{i_k}) \\
 &= \lambda_{i_1} v_{i_1} \otimes \dots \otimes \lambda_{i_k} v_{i_k} \end{align*}$$
so $v_{i_1} \otimes \dots \otimes v_{i_k}$ is an eigenvector of $f^{\otimes k}$ with eigenvalue $\lambda_{i_1} \dots \lambda_{i_k}$, and if $f$ has a basis of eigenvectors then these tensor products form a basis of eigenvectors for $f^{\otimes k}$ acting on $V^{\otimes k}$.
Since the symmetric and exterior powers are quotients of $V^{\otimes k}$ preserved by all $f$ they inherit these eigenvectors. So every monomial in the $v_i$ of degree $k$ is an eigenvector of $f$ acting on the symmetric power $S^k(V)$, and every monomial in the $v_i$ of degree $k$ with distinct factors is an eigenvector of $f$ acting on the exterior power $\Lambda^k(V)$, and if $f$ has a basis of eigenvectors then these are bases of eigenvectors for $S^k(f)$ and $\Lambda^k(f)$ acting on $S^k(V)$ and $\Lambda^k(V)$ respectively.
In particular it follows that $\text{tr}(S^k(f))$ and $\text{tr}(\Lambda^k(f))$ are symmetric polynomials of the eigenvalues of $f$, namely the complete homogeneous symmetric polynomial $h_k$ and the elementary symmetric polynomials $e_k$ respectively. These can be organized into generating functions
$$H(t) = \sum_{k \ge 0} \text{tr}(S^k(f)) t^k = \frac{1}{\det(1 - tf)}$$
$$E(t) = \sum_{k \ge 0} \text{tr}(\Lambda^k(f)) t^k = \det(1 + tf)$$
which immediately gives a nice symmetric function identity given by expanding out the identity $H(t) E(-t) = 1$.
Generally speaking if $S^{\lambda}(V)$ is any Schur functor then the trace $\text{tr}(S^{\lambda}(f))$ must be a symmetric polynomial in the eigenvalues of $f$ and this is the Schur polynomial $s_{\lambda}$.
