Let $V$ be a vector space and $\alpha$ a nilpotent endomorfism (of degree $k$), how can I show that $\alpha(x)+x$ is epic? 
Let $V$ be a vector space and $\alpha$ a nilpotent endomorfism (of degree $k$), how can I show that $\alpha(x)+x$ is epic? (Exercise 770 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know)

If $v\in V$ I want to show that there exists $x\in V$ such that $\alpha(x)+x=v$ so $x=v-\alpha(x)$.
since $\alpha$ is nilpotent I know that $\alpha^d(v)=0$ and so $\alpha^{d+1}(x)+\alpha^{d}(x)=0$.
I also have that $\alpha^{k-1}(x)=\alpha^{k-1}(v)$
But I just prove it when $v=0$ or $\alpha(v)=0$, so please lend me a hand.
Thanks.
 A: Suppose that $x$ exists, then $x=v-\alpha(x)$
$\Rightarrow x=v-\alpha(v-\alpha(x))=v-\alpha(v)+\alpha^2(x)$
$\Rightarrow x=v-\alpha(v)+\alpha^2(v-\alpha(x))=v-\alpha(v)+\alpha^2(v)-\alpha^3(x)$
$\vdots$
'till get $x=v-\alpha(v)+\alpha^2(v)-\alpha^3(v)+\cdots+(-1)^{k-1}\alpha^{k-1}(v)$, since $\alpha^k(x)=0$.
So let $x=v-\alpha(v)+\alpha^2(v)-\alpha^3(v)+\cdots+(-1)^{k-1}\alpha^{k-1}(v)\in V\Rightarrow \alpha(x)=v$
A: The key here is the fact that, for $\alpha$ nilpotent, $I + \alpha$ is in fact invertible; that is, there exists an endomorphism $\beta$ of $V$ such that
$\beta (I + \alpha) = (I + \alpha) \beta = I; \tag{1}$
then we have, for any $v \in V$,
$(I + \alpha)\beta(v) = I(v) = v; \tag{2}$
thus, setting
$x = \beta(v), \tag{3}$
we have
$x + \alpha(x) = (I + \alpha)(x) = (I + \alpha)(\beta(v)) = v, \tag{4}$
as required; $I + \alpha$ is indeed "epic", as the saying goes.
That $\alpha$ nilpotent implies $I + \alpha$ invertible is a nice result; indeed, it is a good result; its mathematical merit, to my mind at least, lying in the generality with which it holds, that is, over any vector space over any field.  In fact, we could take $V$ to be an $R$ module of some ring $R$ and $\alpha$ an $R$-endomorphism, and the result would still hold, even if $R$ were non-commutative; this should be clear upon reading the following.  But for the moment, we'll stick to vector spaces over fields.
We are given that
$\alpha^d = 0 \tag{5}$
for some positive integer $d$.  Assuming $\alpha \ne 0$, we have $d \ge 2$.  The existence of $\beta$ as above depends on the purely algebraic (that is, not relying on convergence or other topological notions) identities
$(I + \alpha)(\sum_0^m  (-1)^i \alpha^i) = I + (-1)^m \alpha^{m + 1};  \tag{6}$
the first few examples of (6) with $m \ge 1$ are
$(I + \alpha)(I - \alpha) = I - \alpha^2; \tag{7}$
$(I + \alpha)(I - \alpha + \alpha^2) = I + \alpha^3; \tag{8}$
$(I + \alpha)(I - \alpha + \alpha^2 - \alpha^3) = I - \alpha^4, \tag{9}$
and so forth.  (6) is easily proved via a simple induction, taking (7)-(9) as base cases; assuming (6) binds for some positive $k \in \Bbb Z$,
$(I + \alpha)(\sum_0^k (-1)^i \alpha^i) = I + (-1)^k \alpha^{k + 1}, \tag{10}$
we have
$(I + \alpha)(\sum_0^{k + 1} (-1)^i \alpha^i) = (I + \alpha)(\sum_0^k (-1)^i \alpha^i) + (I + \alpha)(-1)^{k + 1} \alpha^{k + 1}$
$= I + (-1)^k \alpha^{k + 1} + (I + \alpha)(-1)^{k + 1} \alpha^{k + 1}$
$= I + (-1)^k \alpha^{k + 1} + (-1)^{k + 1}\alpha^{k + 1} + (-1)^{k + 1} \alpha^{k + 2}$
$= I + ((-1)^k + (-1)^{k + 1}) \alpha^{k + 1} + (-1)^{k + 1} \alpha^{k + 2} = I + (-1)^{k + 1}\alpha^{k + 2}, \tag{11}$
completing the demonstration.  It now follows by (5) that
$(I + \alpha)(\sum_0^{d - 1} (-1)^i \alpha^i) = I + (-1)^{d - 1} \alpha^d = I, \tag{12}$
showing that we may take
$(I + \alpha)^{-1} = \beta = \sum_0^{d - 1} (-1)^i \alpha^i; \tag{13}$
then as in (1)-(4) for any $v \in V$ we may find $x = \beta(v) = (I + \alpha)^{-1}(v)$ such that $x + \alpha(x) = (I + \alpha)(x) = v$. 
Epic!!! It's all so EPIC!
