Let $A\in M_n$, if $\operatorname{rank}(A+I)+\operatorname{rank}(A-I)=n$, find an eigenvalue of $A$.

Question

I can just use the formula to get $\operatorname{rank}(A+I)+\operatorname{rank}(A-I)=n\geq\operatorname{rank}(2A)$. But I don’t know how to use this or otherwise to find one eigenvalue of A. Could anyone help please? Thanks.

Answer 1

By the given rank condition and the rank-nullity theorem, we have $\operatorname{nullity}(A+I)+\operatorname{nullity}(A-I)=n$. Therefore, at least one of $\ker(A+I)$ or $\ker(A-I)$ is a nontrivial vector subspace, meaning that at least one of $-1$ or $1$ is an eigenvalue of $A$.

When the characteristic of the underlying field $\mathbb F$ is not $2$, we can do better: as $\operatorname{nullity}(A+I)+\operatorname{nullity}(A-I)=n$ and $\ker(A+I)\cap\ker(A-I)=0$, we have $\mathbb F^n=\ker(A+I)\oplus(A-I)$. Hence $A$ is similar to $-I_k\oplus I_{n-k}$ for some integer $0\le k\le n$.