If $f \circ f = 0 $, show that transformations $f + id_x$ and $f - id_x$ are isomorphisms of $X$

Since $f \in L(X,X)$ and $f \circ f = 0 $, we have $$(id_x + f )\circ (id_x-f)= id_x -f + f - (f\circ f)= id_x$$ and $$(id_x - f )\circ (id_x + f)= id_x + f - f - (f\circ f)= id_x$$

So $id_x + f $ and $id_x - f $ are isomorphisms. Since $f-id_x = -(id_x-f)$, we have that $f-id_x$ is an isomorphism. So, we have that $f+id_x$ and $f-id_x$ are isomorphisms.

$f - id_X$ is an isomorphism iff $-(f - id_X) = id_X - f$ is an isomorphism.

We have $$[(f + id_X) \circ (id_X - f)](x) = ((f +id_X)((id_X - f)(x))) = (f +id_X)(x - f(x)) = (f + id_X)(x) - (f + id_X)(f(x)) = f(x) + x -f(f(x)) - f(x) = x,$$ i.e. $(f + id_X) \circ (id_X - f) = id_X$. Similarly$(id_X - f) \circ (f + id_X = id_X$.

This shows that $f + id_X, id_X - f$ are inverse isomorphisms.

If $X$ is finite-dimensional, it suffices to show that $f \pm \operatorname{id}_X$ is injective, or that its kernel is trivial.

Assume $0 = (f \pm \operatorname{id}_X)(x) = f(x)\pm x$. Hence $f(x) = \mp\, x$. Applying $f$ to this yields $$0 = f^2(x) = \mp\, f(x)$$ so $x = \mp\, f(x) = 0$.

Hence $f \pm \operatorname{id}_X$ is an isomorphism.

If $X$ is not finite-dimensional, we also need to show that $f \pm \operatorname{id}_X$ is surjective. For this, note that for arbitrary $y \in X$ we have

$$(f \pm \operatorname{id}_X)(\pm \,y -f(y)) = \pm\, f(y)-f^2(y) + y \mp\,f(y) = y$$