Invariant subspaces and eigenvalues

Question

During a lecture my professor explained, that if $A\in \mathbb C^{n\times n},B\in \mathbb C^{k\times k}$ and $C\in\mathbb C^{n\times k}$ with $1\le \operatorname{rank}(C)=k\le n\ $ such that $AC=CB \ $it follows that $\operatorname{Im}(AC)\subseteq \operatorname{Im}(C)$ and $\sigma(B)\subseteq \sigma(A)$. Where $\operatorname{Im}$ denotes the image and $\sigma$ the set of eigenvalues.

Sadly no proof was given and I am unable to figure it out myself. I feel like this should be fairly simple to do, yet I struggle even finding out where to begin. Could somebody nudge me in the correct direction? Many thanks in advance.

Answer 1

The first part is easy: $(AC)v=C(Bv)$, hence any vector in the image of $AC$ belongs to the image of $C$.

Now let $\lambda$ be an eigenvalue of $B$, with eigenvector $v\ne0$, so $$ Bv=\lambda v,\quad v\ne0 $$ Apply the assumption $CB=AC$ to get $$ ACv=CBv=C(\lambda v)=\lambda(Cv) $$ You get therefore that $Cv$ is an eigenvector of $A$ relative to $\lambda$ as soon as you prove that $Cv\ne0$. But $\operatorname{rank}C=k\le n$ is equivalent to $C$ having a left inverse.

Answer 2

Let a$\in \sigma(B)$. Let v$\ne 0$ be an eigen vector of B for eigenvalue a. Then Bv=av. Hence CBv=aCv. Hence ACv=aCv. Since C has full rank, Cv$\ne 0$. Hence Cv is an eigenvector of A with corresponding eigenvalue a. So a$\in \sigma (A)$. Hence the result.