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 $1 \Rightarrow 3$?

  • 5
    $\begingroup$ Have you made the observation yet that $T^2=I\iff(T+I)(T-I)\equiv 0$? $\endgroup$
    – David P
    Aug 25, 2020 at 21:39

1 Answer 1


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. $$ T $$


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