Let $A \in M_n (\mathbb{C})$ with $\operatorname{rank} (I_n + A) + \operatorname{rank} (I_n − A) = n$. Show that $A^2 = I_n$

Question

Let $A \in M_n(\mathbb{C})$ with $$\operatorname{rank} (I_n + A) + \operatorname{rank} (I_n − A) = n$$ Show that $A^2 = I_n$. Hint: Use the Jordan canonical form of $A$.

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If I were to use the Jordan canonical form of $A$, am I supposed to show conjugation between the two ranks while assuming that there exists an invertible square matrix?

I'm having trouble understanding the connection between the ranks, identity matrix and the squaring the matrix itself.

Answer 1

Suppose that $J$ is the Jordan canonical form of $A$. Note that:

• $J + \alpha I$ is the Jordan canonical form of $A + \alpha I$ for any $\alpha$
• $A^2 = I$ if and only if $J^2 = I$
• $-J$ is similar to $-A$
• $A$ and $J$ have the same rank
• $I + A$ and $I + J$ have the same rank
• $I - A$ and $I - J$ have the same rank

So, it's enough to show that if $J$ is a matrix in Jordan canonical form, then $$ \operatorname{rank}(I + J) + \operatorname{rank}(I - J) = n $$ Now, which Jordan block matrices satisfy $J^2 = I$?