How can I prove that a linear map $T:V\to V$ such that $T^2 = I$ is equal to the Identity map?

Question

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

I'm not sure where to start, though I know I somehow have to show that $T \circ T$ yields the identity matrix, but there isn't much information given.

Answer 1

Hint: $0 = T^2-I = (T-I)(T+I)$, with $T+I$ invertible (why?).

... or this:

Hint: $0 = T^2-I = (T+I)(T-I)$, with $T+I$ injective (why?).

Answer 2

Note that

$(T + I)(T - I) = T^2 - I; \tag 1$

since

$T^2 = I, \tag 2$

$T^2 - I = 0; \tag 3$

now suppose

$T \ne I; \tag 4$

then

$\exists \; x \in v, \; Tx \ne Ix = x; \tag 5$

therefore,

$w = (T - I) x = Tx - x \ne 0; \tag 6$

thus, by (1) and (3),

$(T + I)w = (T + I)(T - I)x = (T^2 - I)x = 0(x) = 0; \tag 7$

but

$(T + I)w = 0 \Longrightarrow Tw + w = 0 \Longrightarrow Tw = -w, \tag 8$

which asserts that $w \ne 0$ is an eigenvector of $T$ corresponding to eigenvalue $-1$, in contradiction to the hypothesis placed upon $T$. We see then that (5) cannot bind, whence

$\forall x \in v, \; Tx = Ix = x, \tag 9$

as was to be proved.

Answer 3

Assume that the base field is of characteristic not equal to $2$. First, using the hypothesis $T^2=I$ and the fact that $x-1$ and $x+1$ are coprime polynomials, show that $$V=\text{im}(T+I)\oplus \ker(T+I)\,.$$ Then, prove that $\text{im}(T-I)=\ker(T+I)$. Since $-1$ is not an eigenvalue of $T$, this means $\ker(T+I)=0$. Therefore, $\text{im}(T-I)=0$, proving that $T=I$.

Observe that, for any $v\in V$, $$v=\frac{1}{2}(T+I)(v)-\frac{1}{2}(T-I)(v)\,,\tag{*}$$ with $\frac{1}{2}(T+I)(v)\in \text{im}(T+I)$ and $-\frac12(T-I)(v)\in \text{im}(T-I)\subseteq \ker(T+I)$ (since $(T-I)(T+I)=T^2-I=0$). This shows that $$V=\text{im}(T+I)+\ker(T+I)\,.\tag{#}$$ If $v\in \text{im}(T+I)\cap \ker(T+I)$, then $v=(T+I)(u)$ for some $u\in V$, so $$(T+I)^2(u)=(T+I)(v)=0\,.$$ Since $T^2-I=0$, we get $$v=(T+I)(u)=\frac{1}{2}\big((T+I)^2-(T^2-I)\big)(u)=0\,.$$ Thus, the sum (#) is direct.
From (*), we can also see that $V=\text{im}(T-I)+\text{im}(T+I)$ and $\text{im}(T-I)\subseteq \ker(T+I)$. This implies $\text{im}(T-I)=\ker(T+I)$.