If $f \circ f = 0 $, show that transformations $f + id_x$ and $f - id_x$ are isomorphisms of $X$ 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!
 A: 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.  
A: $f - id_X$ 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.
A: 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
$$(f \pm \operatorname{id}_X)(\pm \,y -f(y)) = \pm\, f(y)-f^2(y) + y \mp\,f(y) = y$$
A: if $h\circ g=idx$ then, $g$ is injective and $h$  is surjective .
$f\circ f-idx=idx=(f-idx)\circ(f+idx)=(f+idx)\circ(cf-idx)$ 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..
A: 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 $f\circ f =0$, 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.
