Ok so here's the problem.
$T$ is a nilpotent linear transformation on a finite dimensional vector space. (Let's say $V=\mathbb{R}^{n}$, without loss of generality.)
Fact: $T$ has only $0$ as an eigenvalue and there is a smallest nonzero natural number, $m$, such that $\text{Ker}(T^{m})=V$.
Show that $T$ can be written as an upper triangular matrix, $A$, with $0$'s on the diagonal where $A$ is with respect to the basis derived as follows:
First find a basis for $\text{Ker}(T)$. Then expand that basis to one for $\text{Ker}(T^{2})$. Expand again for a basis of $\text{Ker}(T^{3})$ and so on until you get a basis for $Ker(T^m) = V\ $ [of course m could be smaller than 3].
Just a reminder: the columns of a matrix are the image of the basis vectors. Therefore if our final basis after the process just explained is {$u_1, u_2, …, u_n$}, then the matrix will be $[T(u_1), T(u_2), …, T(u_n)]$.
Please let me know if you can think of anything. This should be pretty simple, but I'm not seeing something.
Best Answer
You were already given some pretty good hints in the question. However, it probably shouldn't have said "find" a basis, because you don't actually need to find it. ${\rm Ker}(T)$ has a basis, let it be $u_1, u_2, \ldots u_k$. Then ...