Let $R$ be a $k-$algebra and $M,N$ two irreducible $R-$modules, isomorphic as vector spaces. If we know that for every $r\in R$ we have the same eigenvalues on $M$ and $N$ (with multiplicities) is it true that $M\simeq N$?
I have two thoughts:
Denoting $St(m)=\{ rm=m \mid r\in R\}$, if we have $St(m)=St(n)$ for some $m\in M, n\in N$, we could find an isomorphism that sends $m$ to $n$ and is extended by $f(rm)=rf(m)$ (every element of $M$ is $rm$ for some $r$ by irreducibility). I cannot prove the existence of such a pair though.
If $R$ was commutative, we could simultaneously diagonalize all the actions according to some basis and take the isomorphism of vector spaces that sends the one basis to the other. This will then be an isomorphism of modules. I cannot extend that to the general case.
Consider the commutative algebra $R=k[x,y]/(x^2,y^2)$ and the modules $R/(x)$ and $R/(y)$.