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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!

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    Is $X$ finite dimensional? – SmileyCraft Jan 04 '19 at 23:19
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    Try to compose them with each other. – Torsten Schoeneberg Jan 04 '19 at 23:21
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    As people illustrated, one way to prove something is an isomorphism is to actually find its inverse, quite explicitly! A deeper example is the inverse to $I-A$ where norm of the matrix $A$ is small. The trick is to replace formally the variable $x$ below with $A$ -- quite unsettling I know! $(1-x)^{-1}=\frac{1}{1-x} = \sum_{i=0}^\infty x^i$ – Behnam Esmayli Jan 04 '19 at 23:50

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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.

Ramiro
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$f - id_X$ 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.

Paul Frost
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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

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

mechanodroid
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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..

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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.

Torsten Schoeneberg
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