$\DeclareMathOperator\Hom{Hom}$Let $G$ be a (pro-)cyclic group (topologically) generated by $\phi$, and let $V,W$ be two finite-dimensional (continuous) $k$-linear semisimple representations of $G$, where $k$ is some algebraically closed (complete) field.


Let $$P\_V(t) = \prod\_{i=1}^n (t - a\_i), P\_W(t) = \prod\_{j=1}^m (t - b\_j)$$
be the characteristic polynomials of $\rho\_V(\phi), \rho\_W(\phi)$ respectively.


I would like to know if
$$\dim\_k \Hom\_G(V, W) = \# \{ (i, j) : a\_i = b\_j \}$$


I know that $\dim\_k \Hom\_G(V, W) = \langle \chi\_V, \chi\_W \rangle$, and that $f \in \Hom\_G(V, W)$ iff $f \circ \rho\_V(\phi) = \rho\_W(\phi) \circ f$. But I don't know how to proceed from there.


(NB: already asked [here](https://math.stackexchange.com/questions/4355067/) two weeks ago, with no success).



Dimension of $\mathrm{Hom}_G(V, W)$ in terms of characteristic polynomial
If $k$ is of characteristic zero or characteristic $p$ and $G$ is a pro-$p'$-group, then yes. In that case, everything in sight is semisimple by Maschke's theorem, the eigenvalues $a\_i$ and $b\_j$ give you the decomposition into irreducibles, and every pair of equal eigenvalues contributes one degree of freedom by Schur's lemma.


Otherwise no, even in the simplest case, because of lack of semiplicity. If $k=\overline{\mathbb{F}\_p}$, $G=C\_p$, $\rho\_V(\phi) = \begin{pmatrix}1&1\\&1\end{pmatrix}$ and $\rho\_W(\phi)=\begin{pmatrix}1&\\&1\end{pmatrix}$, then $a\_1=a\_2=b\_1=b\_2$ so that your conjecture would predict $\dim\_k \operatorname{Hom}\_G(V,W) = 4$, but it is only 2, because $V$ is not semi simple, but $W$ is, and therefore every morphism must factor through $V/\operatorname{rad}(V)$.


