Proving that the direct sum of these two kernels is the vector space itself Let $V$ be a finite dimensional vector space over a field $F$. Let $f: V \to V$ be a linear transformation and suppose $g$ is the identity linear transformation in $V$.
If $f^2 = g$, prove that
$$\textbf{ker}(f-g) \oplus \textbf{ker}(f+g) = V$$
I think the approach here is to prove that sum of the the nullity of $f-g$ and the nullity of $f+g$ is the dimension of $V$.
 A: We need to assume that $\operatorname{char}F\neq 2$, otherwise take $V=F$ and $f=g=\operatorname{id}$. In any case, $f^2=g$ but $f+g=f-g=0$, so the subspaces in question are not complementar.
Now assuming $\operatorname{char}F\neq 2$: Note that for all $x\in V$, $x-f(x)\in\ker(f+g)$, and that $x+f(x)\in\ker(f-g)$. Therefore
$$x=\frac{x+f(x)}{2}+\frac{x-f(x)}{2}\in\ker(f-g)+\ker(f+g).$$
To finish, we need to show that $\ker(f-g)\cap\ker(f+g)=0$. Suppose $x\in\ker(f-g)\cap\ker(f+g)$. Then
$$f(x)-g(x)=(f-g)(x)=0=(f+g)(x)=f(x)+g(x)$$
so $2x=2g(x)=0$. Since $\operatorname{char}F\neq 2$, we obtain $x=0$.
A: You can try and write a vector $v\in V$ as $v=v'+v''$, with $v'\in\ker(f-g)$ and $v''\in\ker(f+g)$.
Suppose we are able to. Then $f(v')-g(v')=0$, so $f(v')=v'$ and, similarly, $f(v'')=-v''$. Then
$$
f(v)=f(v')+f(v'')=v'-v''
$$
so $v+f(v)=2v'$ and $v-f(v)=2v''$. Hence we have little choice for $v'$ and $v''$; we just need to show that
$$
v'=\frac{1}{2}(v+f(v))\in\ker(f-g),
\qquad
v''=\frac{1}{2}(v-f(v))\in\ker(f+g),
$$

 $f(v')-g(v')=f(v+f(v))-(v+f(v))=f(v)+f(f(v))-v-f(v)=0$You can prove similarly that $v''\in\ker(f+g)$.

You end the proof by showing that $\ker(f-g)\cap\ker(f+g)=\{0\}$.
