TITLE: let $A\in M{_n}\ (\mathbb R)$ s.t $A^2=I$ such that $A\neq I$, $A\neq -I$ how prove $-(n-1)\le tr A\le n-1$
QUESTION [5 upvotes]: let $A\in M{_n}\ (\mathbb R)$ s.t $A^2=I$ such that $A\neq I$, $A\neq -I$ how prove $-(n-1)\le tr A\le n-1.$  Thanks in advance

REPLY [2 votes]: The minimal polynomial of $A$ is $(x-1)(x+1)\Rightarrow $ the characteristic polynomial of $A$ is $$f(x)=(x-1)^k(x+1)^{n-k}=x^n+((n-k)x^kx^{n-k-1}-kx^{k-1}x^{n-k})+\ldots \\=x^n+a_{n-1}x^{n-1}+\ldots, \ \text{where} \ 1\leq k\leq n-1.$$
Now $|\text{tr}(A)|=|a_{n-1}|=|(n-k)-k|=|n-2k|$. 
So we have the stronger result that $|\text{tr}(A)|\leq |n-2|$ 
(and $\text{tr}(A)\in\{2-n,4-n,\ldots,n-4,n-2\}$).