Proof: $I_n + V$ is invertible if and only if $V=I_n$

Question

So I know there are some related posts that tackle similar questions, but in my case, the conditions are slightly different. I have been thinking about this question, but I don't know where to begin.

It is given that $V^2 = I_n$

Prove that $(I_n + V)$ is invertible if and only if $V=I_n$

Could somebody help me? Or at least provide me with a hint?

I have been thinking of using determinants, but I do not see how to apply that in this. (Also, I am not allowed to prove it by using eigenvalues)

Thank you!

Answer 1

$0=V^2-I=(V+I)(V-I)$. If $V+I$ is invertible, then $0=V-I$

Answer 2

The if direction is easy. For the other direction, if $I = (I+V)A$, then $V = V(I+V)A = (V+VV)A = (V+I)A = (I+V)A = I$.

Answer 3

$\bullet\;\;\;$Suppose first $I_n=V$.

Then $I_n+V=2I_n$ which is clearly invertible: its inverse is $\frac12I_n$.

$\bullet\;\;\;$Call now $B:=I_n+V$ and suppose it's invertible.

Then $$ B^2=I^2_n+V^2+2V=2(I_n+V)=2B; $$ now multiply both sides by $B^{-1}$ and get $$ B=I_n+V=2I_n $$ from which immediately $V=I_n$.