Splitting a matrix $A \in \mathbb{M}^{n \times n}(\mathbb{C})$by solving $Av = \lambda C v$ for some chosen $C$

Question

If we know a matrix $A \in \mathbb{M}^{n \times n}(\mathbb{C})$ and solve $Av = \lambda v$ where we try to find $\lambda,v$, we can rewrite $A$ in a nice way. What if we choose a matrix $C$ and we try to solve $Av = \lambda C v$, would that give us another nice decomposition?

Answer 1

It gives the same decomposition, because the equation $Av=\lambda Cv $ only gives an eigenvalue equation $Av=\lambda ' v$ if $Cv=\mu v$. Otherwise $w:=Cv$ and $v$ are not collinear, and solving $Av=\lambda w$ is not related to eigenvalues of $A$. So we may suppose that $Cv=\mu v$. Now $$ Av=\lambda Cv=\lambda \mu v=\lambda' v, $$ which is again an eigenvalue equation like $Av=\lambda v$.

Answer 2

If $C$ is invertible then, $Av = \lambda C v$ is equivalent to $C^{-1}\!Av=\lambda v$, so you're just looking and an eigenvector equation for a different matrix $C^{-1}\!A$ rather than for $A$.