Semisimple representation is determined by characteristic polynomials? Let $\rho, \rho' : G \to GL(V)$ be two finite dimensional semisimple representations of a group $G$ (possibly infinite). If the characteristic polynomials of $\rho(g)$ and $\rho'(g)$ are equal for any $g \in G$, can we conclude that $\rho$ is equivalent to $\rho'$ ? I did not find any reference for this. Thank you!
 A: To give a partial answer:
If $\operatorname{char}(k) = 0$ then we can use the following theorem from Algebra (chapter XVII, $\S3$, page 650) by Serge Lang:

Corollary 3.8 (Bourbaki).
  Let $k$ be a field of characteristic $0$.
  Let $R$ be a $k$-algebra, and let $E, F$ be semisimple $R$-modules, finite dimensional over $k$.
  For each $\alpha \in R$ let $\alpha_E$, $\alpha_F$ be the corresponding $k$-endomorphism on $E$ and $F$ respectively.
  Suppose that the traces are equal;
  that is,
  $$
    \operatorname{tr}(\alpha_E)
  = \operatorname{tr}(\alpha_F)
$$
  for all $\alpha \in R$.
  Then $E$ is isomorphic to $F$ as $R$-module.

We can apply this to the group algebra $R = k[G]$;
let’s denote the extensions of $\rho, \rho'$ to $k[G]$ also by $\rho, \rho'$.
Because $\rho(g), \rho'(g)$ have the same characteristic polynomial for every $g \in G$ it follows that $\operatorname{tr}(\rho(g)) = \operatorname{tr}(\rho'(g))$ for every $g \in G$, and therefore that $\operatorname{tr}(\rho(\alpha)) = \operatorname{tr}(\rho'(\alpha))$ for every $\alpha \in k[G]$.
It then follows from the above theorem that the two representations are isomorphic as $k[G]$-modules and thus as representations.
A: Edited 05/08/2022 There is a much better proof of the full theorem: see comments in Characteristic polynomials determine semisimple representation: possible counterexample
This is a very late answer, but since this question is very important, and I can't find the answer elsewhere, so I decided to write it.
First, we will translate this question into the question of semisimple modules over $k$-algebras.
$\textbf{Lemma 1}$ Let $A$ be an algebra over a field $k$, and $S$ be a $k$-basis of $A$. For any two finite-dimensional $A$-modules $V,W$, if $\text{char}_{V}(a)=\text{char}_{W}(a)$ for $a\in S$, then $\text{char}_{V}(a)=\text{char}_{W}(a)$ for $a\in A$.
$\textbf{proof. }$ Coefficients of characteristic polynomials are traces of exterior powers. Hence this follows from the $k$-linearity of traces.
By this lemma, for any representations $\rho_1, \rho_2$ of $G$, $\text{char}_{\rho_1}(g)=\text{char}_{\rho_2}(g)$ for $g\in G$ implies same characteristic polynomial for all elements of $k[G]$. So we can translate the problem into the following theorem.
$\textbf{Theorem 1}$ Let $A$ be a finite-dimensional algebra over a field $k$. Assume $k$ is perfect, or $A=k[G]$ for some finite group $G$. Let $V,W$ be two semisimple $A$-modules. If $\text{char}_{V}(a)=\text{char}_{W}(a)$ for every $a\in A$, then $V\simeq W$.
This implies our desired conclusion.
$\textbf{Corollary 1}$ Assume $G$ is finite or $k$ is perfect. If two semisimple representations $\rho_1,\rho_2$ (acts on $k$-vector spaces $V_1,V_2$) of $G$ over $k$ has same characteristic polynomials, then they are isomorphic.
$\textbf{proof. }$ If $G$ is finite, this directly follows from Theorem 1 with $A=k[G]$. If $k$ is perfect, let $A$ be the image of $k[G]$ on $\text{End}(V_1\oplus V_2)$. Then $A$ is a finite-dimensional $k$-algebra. Then, by Theorem 1, $V_1, V_2$ are isomorphic as $A$-module. Hence they are isomorphic as $k$-modules.
$\textbf{Theorem 2 [Density Theorem]}$ Let $k$ be an algebraically closed field and $A$ be a finite-dimensional algebra over $k$. Let $V_1,\cdots, V_n$ be pairwise nonisomorphic simple $A$-modules. Then the natural map $A\to \oplus_{i}\text{End}_k(V_i)$ is surjective.
This is a very standard theorem. For the proof, see Theorem 2.5 of this note.
$\textbf{Theorem 3 [Linear independence of characters]}$ Let $k$ be an algebraically closed field and $A$ be a finite-dimensional algebra over $k$. Let $V_1,\cdots, V_n$ be pairwise nonisomorphic simple $A$-modules and $\chi_{V_i}$ be their characters. Then $\chi_{V_i}$ are linearly independent over $k$.
$\textbf{proof. }$ Assume $\sum_i c_i\chi_{V_i}(x)=0$ for some $c_i\in k$ and for all $x\in A$. By Theorem 2, for any $k$ there exist an element $a_k\in A$ such that $a_k$ acts on $V_k$ as $1$ and as $0$ on $V_j$, $j\ne k$. Then
$$0=\sum_i c_i\chi_{V_i}(a_kx)=c_k\chi_{V_k}(x).$$
Now, since $A\to \text{End}_k(V)$ is surjective, we have $c_k=0$ for all $k$.
$\textbf{proof of Theorem 1 for algebraically closed fields.}$ Assume $k$ is algebraically closed. Let $V=\oplus_{i}V_i^{n_i}$ and $W=\oplus_{j}W_j^{m_j}$. By the assumption, $\sum_i n_i\chi_{V_i}=\sum_j m_j\chi_{W_j}$. If $\text{char }k=0$, then characters of simple modules are linearly independent over $\mathbb{Q}$, hence this implies $V\simeq W$.
Now assume $\text{char }k=p$. Assme $V\not\simeq W$. By deleting common simple factors, we may assume that $V_i\not\simeq W_j$ for any pair of $(i,j)$. Since $\chi_{V_i}$ and $\chi_{W_j}$ are linearly independent over $k$, we have $p|n_i, m_j$ for every $i,j$. Define $n_i'=n_i/p$ and $m_p'=m_j/p$ and let $V'=\oplus_{i}V_i^{n_i'}$ and $W'=\oplus_{j}W_j^{m_j'}$. Then, by direct calculation, $\text{char}_{V'}(a)=\text{char}_{W'}(a)$. If $V'\simeq W'$, then $V\simeq W$, hence we have $V'\not\simeq W'$. We can infinitely repeat this procedure, hence it contradicts to finite dimensionality.
To prove this for the general field, we need to prove the following theorem.
$\textbf{Theorem 4}$ Assume $k$ is perfect, or $A=k[G]$ for some finite group $G$. If $V$ is finite-dimensional semisimple $A$-algebra, then $V_{\bar{k}}$ is also a semisimple $A_{\bar{k}}$-algebra.
I will use some facts from T. Y. Lam's "A first course in noncommutative rings" to prove this theorem. For any ring $R$, let $\text{rad }R$ be its Jacobson radical.
$\textbf{Proposition 1 [direct corollary of Theorem 4.14 of Lam]}$ Let $A$ be a finite-dimensional $k$-algebra. Then $A/\text{rad }A$ is a semisimple $k$-algebra.
$\textbf{Corollary 2}$ An $A$-module $V$ is semisimple if and only if $(\text{rad }A)V=0$
$\textbf{proof. }$ "only if" part is just the definition of Jacobson radical. If $(\text{rad }A)V=0$, then $V$ is an $A/\text{rad }A$-module, which is a semisimple algebra. Hence $V$ is semisimple.
$\textbf{Proposition 2 [Theorem 5.17 of Lam]}$ If $K/k$ is a separable algebraic extension, then $(\text{rad }A)\otimes_k K=\text{rad }A\otimes_k K$.
$\textbf{Proposition 3 [Exercise 6.8 of Lam]}$ For any field extension $K/k$ and a finite group $G$, $\text{rad }K[G]=(\text{rad }k[G])\otimes K$.
Now, Theorem 4 easily follows from Corollary 2 and Proposition 2,3. Finally, we will prove Theorem 1 for the general field. By Theorem 4, $V_{\bar{k}}$ and $W_{\bar{k}}$ are semisimple $A_{\bar{k}}$-modules. By deleting common simple factors, we may assume that $V$ and $W$ share no simple factors. Then $\text{Hom}_A(V,W)=0$, so $\text{Hom}_{A_{\bar{k}}}(V_{\bar{k}},W_{\bar{k}})=0$. Since $V_{\bar{k}}$ and $W_{\bar{k}}$ are semisimple, this means they has no isomorphic simple factors. However, they have same characteristic polynomials (since $V$ and $W$ do, and remember Lemma 1), they are isomorphic. Hence $V_{\bar{k}}=W_{\bar{k}}=0$. So we must have (since we deleted common simple factors) $V\simeq W$.
Now, this is the problem: Is Corollary 1 still hold when $k$ is non-perfect? Does it at least hold for the image of $k[G]$ for some group $G$? Theorem 4 certainly not. $\mathbb{F}_2(t)[x,1/x]/(x^2-t)$ is a simple $\mathbb{F}_2(t)[x,1/x]=\mathbb{F}_2(t)[\mathbb{Z}]$-module, but it is not semisimple over $\mathbb{F}_2(t^{\frac{1}{2}})$. However, I cannot find any counterexample of Theorem 1 or corollary 1. If anyone knows a counterexample, please let me know.
