I have come across an oddity in the solution to a problem and I am unsure with how to proceed.
For an $n\times n$ (symmetric and real-valued, usually poorly conditioned and low-rank) matrix $A$, and an $n\times m$ matrix $B$ (with $m \leq n$), and a scalar $c \neq 0$, the solution comes where:$$AB = cB$$ If $B$ were a vector, this would be a straightforward eigenvalue problem. However, $B$ is in this case a tall skinny matrix.
My intuition says that each column of $B$ must be an eigenvector of $M$ associated with the eigenvalue $c$. But as far as I can tell the only repeated eigenvalue of $M$ is $0$. Is there a solution to this problem?
It is indeed the case that $AB = cB$ means that every column of $B$ is an eigenvector of $A$ associated with eigenvalue $c$. Indeed, if $B_1,\dots,B_n$ denote the columns of $B$, then we have $$ AB = cB \implies A[B_1 \ \ \cdots \ \ B_n] = c[B_1 \ \ \cdots \ \ B_n] \implies \\ [AB_1 \ \ \cdots \ \ AB_n] = [cB_1 \ \ \cdots \ \ cB_n]. $$ The column span of $B$ must be a subspace of the eigenspace associated with $c$, so the rank of $B$ is at most equal to the multiplicity of the eigenvalue $c$.