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

[zenterix]

Homework Statement

Suppose $T\in\mathcal{L}(V)$ and $T^2=I$ and $-1$ is not an eigenvalue of $T$. Prove that $T=I$

Relevant Equations

Since $T^2=I$ then $T_2-I=0$ is the zero operator. Then $(T+I)(T-I)=0$.

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.

 

[PeroK]

Your solution looks fine.

 

[martinbn]

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 $T-I=0$.