Forming a matrix from a nilpotent operator 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.
 A: 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 ...  
A: This is a response to user11314's comment to Robert Israel's answer.
Let $n_1 = \dim(\mathrm{ker}(T))$, let $n_1+n_2=\dim(\mathrm{ker}(T^2))$, and so on, until $\dim(\mathrm{ker}(T^m)) = n_1+n_2+\cdots+n_m = \dim(V)$.
Let $u_1,\ldots,u_{n_1}$ be the basis you find for $\mathrm{ker}(T)$. Then $u_{n_1+1},\ldots,u_{n_1+n_2}$ the extension for a basis of $\mathrm{ker}(T^2)$.
The key observation is that if $\mathbf{v}\in\mathrm{ker}(T^k)$, then $T(\mathbf{v})\in\mathrm{ker}(T^{k-1})$.
So, think about the matrix you get from this basis. The first $n_1$ columns are $0$, since $u_1,\ldots,u_{n_1}$ are all in the kernel.
The next $n_2$ columns correspond to the images of $u_{n_1+1},\ldots,u_{n_1+n_2}$; each of these vectors lie in $\mathrm{ker}(T^2)$ but not in $\mathrm{ker}(T)$, so their images lie in $\mathrm{ker}(T)$ and therefore are expressed using only the vectors $u_1,\ldots,u_{n_1}$; so these columns have $0$s below the $n_1$st row, i.e., are zeros at and below the diagonal at least.
The next $n_3$ columns correspond to images of $u_{n_1+n_2+1}$ through $u_{n_1+n_2+n_3}$. Since these images lie in $\mathrm{ker}(T^2)$, they are expressed using $u_1,\ldots,u_{n_1+n_2}$, so the nonzero entries in these columns lie above the main diagonal.
And so on.
