(If $A\mathbb{x}=\lambda\mathbb{x}$ and $B=TAT^{-1}$ for $T$ invertible then $\mathbb{y}=B\mathbb{x}$ satisfies $B\mathbb{y}=\lambda\mathbb{y}$ .) I have oversimplified because actually the geometry of the solution depends on the signature of $A$, an invariant including $k$ but also including the number of zero eigenvalues, counting multiplicity or viewed as the multiset of eigenvalues, called the