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I have a problem with this task:

The linear transformation $f in L(X,X)$ has property $f circ f = 0 $
Show that transformations $f + id_x$ and $f - id_x$ are isomorphisms of $X$ space with itself


If I need to be honest I have no idea how to prove this. I was tryinging something like that:


If $f + id_x$ is isomorphism it must be both injective and surjective.

Ok, but $id_x$ is injective and surjective.

I thought to show that f is surjective but unfortunately sum of two surjectives can give me something what is not surjective...

Maybe the key is in $f circ f = 0 $?
Thanks for your time!

$f - id_X$ 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.

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

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.

if $hcirc g=idx$ then, $g$ is injective and $h$ is surjective .

$fcirc f-idx=idx=(f-idx)circ(f+idx)=(f+idx)circ(cf-idx)$ and this means that:
-$f+idx$ is injective and $f-idx$ is surjective
-$f-idx$ is surjective and $f+idx$ is injective

And finally you got the answer..

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

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

As all the other answers have amended my comment by actually going through the computation, I just want to add two levels of generalisations of the result which I see:

i) We only need the abelian group structure on $X$. In other words, with the same proof we get the more general result:

If $A$ is an abelian group and $f in End(A)$ satisfies $fcirc f =0$, then $pm id_A pm f$ are automorphisms of $A$.

ii) Now the $End(A)$ in part i is a unital (but not necessarily commutative) ring $R$ with addition as addition and composition as multiplication, $id_A =1_R$. From this perspective, the result is just a special case of

If $u in R^times$, $n in R$ is nilpotent, and $u$ and $n$ commute with each other, then $u+n in R^times$.

which is proved by a geometric series argument as suggested in Behnam Esmayli's comment. In our special case $n=pm f$ with $n^2=0$, of course the inverse of $u+n$ is just $u^{-1}(1-n)$. See Units and Nilpotents and its many duplicates for a general discussion.