Rank of a linear combination of an involutory matrix and the identity matrix

Question

Let $A$ be a square $n \times n$ matrix which satisfies $A^2=E$, where $E$ is an $n \times n$ identity matrix. Prove that $$\mbox{Rank} (A+E) + \mbox{Rank} (A-E)=n$$

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Answer 1

Choose $y\in\text{Col}(A+E)\cap\text{Col}(A-E)$. This means $$(A+E)x_1=y$$ $$(A-E)x_2=y$$ for some $x_1,x_2\in \mathbb{R}^n$. Apply $A$ to both sides of the first equation on the left and get $Ay=y$. Doing the same on the second equation yields $Ay=-y$. By transitivity we have $y=-y$ implying $y=0$. This means $$\text{Col}(A+E)\cap \text{Col}(A-E)=\{0\}$$

Now observe for any $x\in \mathbb{R}^n$ we have $$x=(A+E)\Big(\frac{x}{2}\Big)+(A-E)\Big(-\frac{x}{2}\Big)$$ which shows $$\mathbb{R}^n=\text{Col}(A+E)\oplus\text{Col}(A-E)$$ The result follows.

Answer 2

where $\text{char}\big(\mathbb F\big) \neq 2$, we have
$n\leq \mbox{Rank}\Big(A+E\Big) + \mbox{Rank} \Big(A-E\Big) \leq n$, because $\mbox{Rank} \Big((A+E)+ (A-E)\Big) \leq \mbox{Rank} \Big(A+E\Big) + \mbox{Rank} \Big(A-E\Big)\leq \mbox{Rank} \Big((A+E)(A-E)\Big)+n $

i.) $(A+E)+ (A-E)=2A$ and $A$ is invertible so the lower bound is $n$
ii.) sub-additivity of rank
iii.) Sylvester's Rank Inequality
iv.) $(A+E)(A-E)=A^2-A+A-I=I-I=\mathbf 0$, so the upper bound is $n$

note $2A=\mathbf 0$ in characteristic 2 so the lower bound loosens completely there, and e.g. $A:=E$ meets said lower bound with equality.