Someone gave me the following problem: Let $E$ be a complex vector space of dimension $n\ge 2$.
Let $u_1,\cdots,u_r$ be $r$ endomorphisms of $E$ such that for all $k$ $$u_k^2=-\operatorname{id},$$ and, if $k\neq l$, $$u_k\circ u_l=-u_l\circ u_k.$$ Show that $u_k$ is an automorphism for each $k$ and that $n$ must be even.
My problem with this is the following: The second relation gives that $$\det(u_k)\det(u_l)=-\det(u_l)\det(u_k),$$ hence $\det(u_k)\det(u_l)=0$ hence one of them is with determinat zero and so it can not be an automorphism, is my reasoning correct ? Thank you for your help!
You are given that $u_k^2=-\operatorname{id}$ where $-\operatorname{id}$ is an automorphism, so $u_k$ must be an automorphism.
You are given that $u_k\circ u_l=-u_l\circ u_k$, so if we set $n:=\dim E$ then $$\det(u_k)\det(u_l)=\det(u_k\circ u_l)=\det(-u_l\circ u_k)=(-1)^n\det(u_l)\det(u_k),$$ where $\det(u_l)\neq0$ and $\det(u_k)\neq0$ because they are automorphisms, so $(-1)^n=1$ so $n$ is even