Let $\mathbf{A} \in \mathbb{C}^{n \times n}$ and $\mathbf{X} \in \mathbb{C}^{n \times p}$ with $Rank(\mathbf{X})=p$ for $1 \leq p < n$. I want to show that if $\mathbf{A}\mathbf{X} = \mathbf{X}\mathbf{M}$, then the eigenvalues of $\mathbf{M}$ are eigenvalues of $\mathbf{A}$.
I have been trying to use QR and Schur decompositions to get some sort of similarity transformation that would prove this statement. However, I am having trouble with this. I also am not sure where $\mathbf{X}$ having full column rank fits into this.
If $\lambda$ is an eigenvalue of $M$ and $Y \neq 0$ is an eigenvector corresponding to $\lambda$, then $$A(XY) = (AX)Y = (XM)Y = X(MY) = \lambda(XY);$$
Since $X$ has full rank, $XY \neq 0$, hence $\lambda$ is an eigenvalue of $A$.