Exercise on linear maps with a nilpotent one sorry for asking to help me with this trivial problem. Unfortunately I'm in a very bad shape with linear algebra, being this the fourth exercise I'm not able to solve.
I need some suggestion.
$f,g:V\to V$ linear maps, with $f$ being nilpotent and
$fg-gf=f$
Prove that $\ker(f)$ is invariant for $g$.
I think I have to prove that $g(\ker(f))\subseteq\ker(f)$.
I know that if $f$ is nilpotent, then there exists a $k>1$ for which $f^k=0$, and, moreover $f$ has the only eigenvalue $\lambda=0$.
This problem shouldn't be hard to solve. In your opinion, what do I have to revise to learn making this exercises? Maybe I should start with something simpler?
Thank you.

Let $x$ be an element of $\ker(f)$. Then
$$
0 = f(x) = (fg - gf)(x)
$$
implies (by linearity)
$$
f(g(x)) = g(f(x)) = g(0) = 0
$$
i.e. $g(x) \in \ker(f)$. Since the choice of $x$ was arbitrary, this implies that $g(\ker(f)) \subseteq \ker(f)$.

All you need to know to solve this is what the kernel of a linear map is and that a linear map always sends $0$ to $0$. Then you observe that $f$ and $g$ commute (i.e. $f \circ g = g \circ f$) on $\ker(f)$.
