Ok so here's the problem.
T is a nilpotent linear transformation on a finite dimensional vector space(let's say V=R^n WLOG) into itself. fact: T has only 0 as an eigenvalue and there is a smallest nonzero natural number, m, such that 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 following basis: First find a basis for Ker(T). Then expand that basis to one for Ker(T^2). Expand again for a basis of 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.
