Operator T, ##T^2=I##, -1 not an eigenvalue of T, prove ##T=I##.

Now, for ##v\in V##, ##(T+I)v=0\implies Tv=-v##. That is, the null space of ##T+I## is formed by eigenvectors of ##T## of eigenvalue ##-1##.

By assumption, there are no such eigenvectors (since ##-1## is not an eigenvalue of ##T##).

Hence, if ##(T-I)v \neq 0## then ##(T+I)(T-I)v\neq 0##.

Thus, For ##(T+I)(T-I)=0## we need to have ##(T-I)v=0## for all ##v \in V##.

##(T-I)v=Tv-v=0\implies Tv=v##

Therefore, it must be that ##T=I##.

This is the solution I came up with.

For comparison, here is another solution. My solution differs from this solution, I just want to make sure my reasoning is correct since my proof seems simpler than the linked solution.

Your solution looks fine.

You can make it a bit shorter. If the null space of ##T+I## contains only the zero vector. This operator is invertable so ##(T+I)(T-I)=0## implies ##T-I=0##.