let V be a vector space over a field $F$and let $\alpha \in End(V)$ be nilpotent, having index of nilpotence $k>0.$Show that $\sigma_{1} + \alpha \in Aut(V).$
where $\sigma_{c}$ is defined as $\sigma_{c}(v) = cv.$
My thoughts:
By definition of $Aut(V),$ I should show that $\sigma_{1} + \alpha $ is a linear transformation, 1-1 and onto.
For the linear transformation part, I have shown that:$\forall x,y \in V$ and $\forall \gamma \in F, (\sigma_{1} + \alpha)(x + \gamma y) = (\sigma_{1} + \alpha )(x) + \gamma (\sigma_{1} + \alpha)(y).$ I did not use the hypothesis that $\alpha$ is nilpotent in any step of this proof .... or should this step include the usage of this hypothesis?
Second for proving 1-1, I stopped at $x + \alpha(x) = y + \alpha(y)$ but then what? how can I show that $x = y$?
For the onto part, I do not know why if we are given an element $v \in V,$ why we are sure that there exists an element $w \in V$ such that $(\sigma_{1} + \alpha)(w) = v$, could anyone help me in proving this please?
Recall this basic algebraic identity when $ab=ba$:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}).$$
And $\sigma_1$ commutes with everything, so we have
$$\sigma_1^k-(-\alpha)^k=(\sigma_1+\alpha)\big(\sigma_1^{k-1}-\sigma_1^{k-2}\alpha+\sigma_1^{k-3}\alpha^2-\cdots+\sigma_1(-\alpha)^{k-2}+(-\alpha)^{k-1}\big)$$
which simplifies to
$$\sigma_1-0=\sigma_1=(\sigma_1+\alpha)\Big(\sigma_1-\alpha+\alpha^2-\cdots+(-\alpha)^{k-2}+(-\alpha)^{k-1}\Big).$$
Therefore, $\sigma_1+\alpha$ has an inverse, which is the factor on the right. Invertible means both one-to-one and onto.