# How $T^2 = \text{id}_V$ implies these other properties?

Let $$V$$ be a vector space over $$\mathbb{R}$$ and $$T$$ a linear operator on $$V$$. Prove that the following conditions are equivalent:

1. $$T^{2}= \text{id}_V$$,
2. $$V$$ is the direct sum of the null space (kernel) of $$T- \text{id}_V$$ and the null space of $$T+ \text{id}_V$$,
3. there exist two subspaces, $$W$$ and $$X$$, of $$V$$ such than $$V=W \oplus X$$ and $$T(w+x)=w-x$$ for all $$w \in W$$ and all $$x \in X$$.

I've already proved that $$2 \Leftrightarrow 3$$ and $$3 \Rightarrow 1$$. How can I prove that $$1 \Rightarrow 2$$ or $$1 \Rightarrow 3$$?

• Have you made the observation yet that $T^2=I\iff(T+I)(T-I)\equiv 0$? Aug 25, 2020 at 21:39

You have already done most of the work. For $$1\Rightarrow 2$$, assume $$T^2=id$$. Given any vector $$v\in V$$, define $$w=T v-v$$, $$x=T v+v$$. Then, $$Tw=T^2 v- T v=v - Tv=-w$$ so $$w\in Ker(T+id)$$. Also, $$Tx=T^2 v+ T v=v + Tv=x$$ and $$x\in Ker(T-id)$$. But $$v=\frac{1}{2} x-\frac{1}{2} w$$ so $$V=Ker(T+id)+Ker(T-id)$$.
Finally, to see that the sum is direct, consider $$v\in Ker(T+id)\cap Ker(T-id)$$. Then, both $$Tv=v$$ and $$Tv=-v$$ hold, which added give $$2Tv=0$$ and also $$T(2Tv)=0$$. But since $$T^2=id$$, we have $$2T^2v=2v=0$$, or $$v=0$$, so the intersection of the kernels is trivial. $$T$$