Let $V$ be a vector space and $\alpha$ a nilpotent endomorfism (of degree $k$), how can I show that $\alpha(x)+x$ is epic?

Question

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.

Answer 1

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 $x=v-\alpha(v)+\alpha^2(v)-\alpha^3(v)+\cdots+(-1)^{k-1}\alpha^{k-1}(v)\in V\Rightarrow \alpha(x)=v$

Answer 2

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!