Show that if $A^2$= $I_n$ (Identity Matrix) then the eigenvalue of A, $\lambda$ is $1$ or $ -1$ The question continues: Suppose that F= $\mathbb Q$ (rational number), $\mathbb R$ (real number) or $\mathbb C$ (complex number). Show that if $A^2$= $I_n$ (Identity Matrix) then the eigenvalue of $A$, $\lambda$, is 1 or -1. Show that ker(LIn+A)=E-1(A), im(LIn+A)=E1(A). Similarly, show that ker(LIn-A)=E1(A), im(LIn-A)=E-1(A). How to solve this? Since A will not necessarily be In or -In, then how can I start?

If $\lambda$ is a eigenvalue of $A$ then exist $v\
ot =0$ such that $Av=\lambda v$, then $A^2v=\lambda Av=\lambda \lambda v=\lambda^2 v$, using that $A^2=I_n$, $v=\lambda^2 v$. Then, $\lambda^2=1$, since $v\
ot=0$.

Remark then that $w \in E_{-1}(A)$ iff $Aw=-w$ iff $(A+I_n)w=0$ iff $w\in Ker(A+I_n)$. Too $y\in Im (A+I_n)$ iff exists $x$ such that $(A+I_n)x=y$ then $$ 0=(A^2-I_n)x=(A-I_n)(A+I_n)x=(A-I_n)y $$ i.e., $Ay=y$, then $y\in E_1(A)$. If $y\in E_1(A)$, then $Ay=y$. Let $x=\frac{1}{2}y$, then $(A+I_n)x=y$, then $y\in Im (A+I_n)$. Thus $Im (A+I_n)=E_1(A)$.

For the other case, the previous result applies to $B=-A$.