# Suppose $T$ is an operator on $V$ and $T^2 = I$ and $-1$ is not an eigenvalue of $T$. Prove that $T = I$.

My attempt:

Since $$-1$$ is not an eigenvalue of $$T\implies T+I$$ is invertible, that is $$(T+I)^{-1}$$ exists. Now $$(T+I)(T-I) = T^2 - I \implies (T+I)(T-I) = 0$$.

After applying $$(T+I)^{-1 }$$ on left and right hand side of the previous equation we get: $$(T+I)^{-1} (T+I) (T-I) = (T+I)^{-1}(0) \implies T-I = 0 \implies T = I$$.

• yes, that's correct Commented May 5, 2022 at 20:57
• Your proof looks fine to me. Here is another way to see it: Suppose $T(\vec{v}) \ne \vec{v}$ for some vector $\vec{v}$, then $T(\vec{v}) - \vec{v} \ne 0$ is a eigenvector with eigenvalue $-1$.
– Nate
Commented May 5, 2022 at 21:01
• Another way: since $T^2-I=0$, the minimal polynomial of $T$ divides $x^2-1=(x+1)(x-1)$. But the roots of the minimal polynomial are the eigenvalues of $T$; since $-1$ is not an eigenvalue, the minimal polynomial must actually divide $x-1$, so that $T-I=0$. This illustrates a general method—but that being said, in this example I like your solution better! Commented May 5, 2022 at 21:09
• If $(T-I)v \neq 0$ for some $v$, then $T+I$ injective implies $(T+I)((T-I)v) \neq 0$. By assumption, that can't happen, so $\forall v~((T-I)v=0)$ Commented May 5, 2022 at 22:57
• @DarbyBond I prefer to argue that way: as $(T+I)(T-I)v=0,$ then $(T-I)v=0.$ Since $v$ is arbitrary $T-I=0.$ Commented May 6, 2022 at 2:26