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!

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.
