If $A$ is not necessarily an invertible $n\times n$ matrix, show that $S^{-1}(v)$ where $v$ is an eigenvector of $A$ and $SAS^{-1}$?

Question

More specifically, how can we prove that if $S$ is invertible but $A$ isn't necessarily invertible and $A$ and $SAS^{-1}$ share eigenvalue lambda that the eigenvector related to lambda for $SAS^{-1}$ will be $S^{-1}(v)$?

I've currently worked out that since they share an eigenvalue, $A$ could be similar to itself – but I don't see how to proceed with the proof from there.

Answer 1

It should be $Sv$ instead: If $v$ is an eigenvector for $A$ with eigenvalue $\lambda$, meaning that $Av=\lambda v$, then $$ SAS^{-1}(Sv)=SAv=S(\lambda v)=\lambda Sv $$ so $Sv$ is an eigenvector for $SAS^{-1}$ with eigenvalue $\lambda$.