Let $V$ be an n-dimensional vector space and define the operators $T, \space G:V\rightarrow V$ over an arbitrary field $\mathbb{F}$, with char($\mathbb{F})=0$. Suppose that $T^n=0$, $\text{null}(T) = 1$, and $GT- TG = T.$ Prove that the eigenvalues of $G$ are of the form $\alpha,$ $\alpha -1, \alpha-2, \dots, \alpha-(n-1)$, for some $\alpha \in \mathbb{F}$.
My attempt:
By assumption, $\text{null}(T)=1$, so there exists a nonzero vector $v \in \text{ker}(T)$. Clearly, $0= T(v)=(GT- TG)(v) \Rightarrow T(G(v))=0 \Rightarrow G(v) \in \text{ker}(T)$ which implies that $\text{ker}(T)$ is $G$-invariant.
Let $u$ be an eigenvector of $G$, then $G(u) = \lambda u$ for some $\lambda \in \mathbb{F}$ and we have that $(GT - TG)(u) = T(u) \Rightarrow G(T(u)) = (\lambda+1)T(u)$.
Therefore, if $u$ is eigenvector of $G$, with an eigenvalue $\lambda$, then $T(u)$ is an eigenvector with eigenvalue $(\lambda+1)$.
But I don't know what I will do with this..
Hint: You're nearly there. Slight correction: suppose that $u$ is an eigenvector of $G$ with eigenvalue $\lambda$. Then, as you calculated, we find that $$ G(T(u)) = (\lambda+1)T(u) $$ That is, we have two possibilities: either $T(u)$ is an eigenvector of $G$ associated with $\lambda+1$ or $T(u) = 0$, which is to say that $T(u) \in \ker G$. Because $\operatorname{null}(T) = 1$, we can deduce that $T(u) = 0$ will happen for exactly one eigenvector of $G$.