Finding eigenvalues of Sym$^k(\varphi)$, with $\varphi$ an operator from an $n$-dimensional vector space $V$ to itself today I'm interested in solving a linear algebra problem concerning the symetric application.
Problem is: I work on an $n$-dimensional vector space $V$ with an operator $\varphi : V \to V$ having a completely decomposable characteristic polynomial. Considering $\{\mu_i\}$ to be the eigenvalues of $\varphi$, I would like to show that the eigenvalues of Sym$^k(\varphi)$ are given by $\{\mu_{i_1}...\mu_{i_k}\}$, for $1 \leq i_1 \leq \cdots \leq i_k \leq n$.
From there I have two questions concerning the data, first: is Sym the application: $\underbrace{V \otimes \cdots \otimes V}_{n \text{ times}} \to \underbrace{V \otimes \cdots \otimes V}_{n \text{ times}}$ such that Sym$(v_1 \otimes \cdots \otimes v_n) = \frac{1}{n!}\sum\limits_{\sigma \in S_n}v_{\sigma(1)} \cdots v_{\sigma(n)}$? Else what is it?
Second: I assume that a completely decomposable characteristic polynomial is such that it has $n$ (since dim($V) = n$) eigenvalues, all of algebraic multiplicity 1, but is that right? Because I never saw this expression before and I didn't find its equivalent in a book or on mathstack.
As you can guess, I can't move on the real question without this info, so any clue about this is welcome!
Have a nice day
 A: When $\varphi:V\rightarrow V$ is a linear map, it induces a linear map $\varphi^{\otimes k}:V^{\otimes k}\rightarrow V^{\otimes k}$ by setting $$
\varphi(v_1 \otimes v_2\otimes \dots\otimes v_k)=\varphi(v_1)\otimes\dots\otimes\varphi(v_k)
$$ on rank 1 tensors and extending linearly. Consider the subspace $S^k(V)$ of symmetric $k$-tensors. You may show that $\varphi^{\otimes k}(S^k(V))\subset S^k(V)$, i.e., $\varphi^{\otimes k}$ maps $S^k(V)$ to itself. This map is what you denote by $Sym^k(\varphi)=\varphi^{\otimes k}_{\mid S^k(V)}$.
Now say $\varphi$ is a diagonalizible matrix with eigenvalues $\mu_1,\mu_2,\dots,\mu_n$ and eigenvectors $v_1,v_2,\dots,v_n$. Now, set $$
v_{i_1}v_{i_2}\dots v_{i_k}=\sum_{\sigma\in S_k} v_{i_{\sigma(1)}}\otimes v_{i_{\sigma(2)}}\otimes\dots\otimes v_{i_{\sigma(k)}}
$$ be the symmetric product. All symmetric products of this form gives us a basis for $S^k(V)$. Observe that $$
\begin{split}
Sym^k(\varphi)(v_{i_1}v_{i_2}\dots v_{i_k}) &= Sym^k(\varphi)\big(\sum_{\sigma\in S_k}v_{i_{\sigma(1)}}\otimes\dots\otimes v_{i_{\sigma(k)}}\big)\\
&=\sum_{\sigma\in S_k}\varphi(v_{i_{\sigma(1)}})\otimes\dots\otimes \varphi(v_{i_{\sigma(k)}})\\
&=\mu_{i_1}\mu_{i_2}\dots\mu_{i_k}\sum_{\sigma\in S_k}v_{i_{\sigma(1)}}\otimes\dots\otimes v_{i_{\sigma(k)}}\\
&=(\mu_{i_1}\mu_{i_2}\dots\mu_{i_k}) v_{i_1}v_{i_2}\dots v_{i_k}.
\end{split}
$$ This shows the result for diagonalizible $\varphi$. However, the set of diagonalizible matrices is dense in $End(V)$, so the result also holds for all $\varphi$.
