How to prove that a nilpotent operator has a basis representation that is strictly upper triangular?

Solution 1:

First of all, it is not necessarily the case that the elements $T^j v$ for $j=0,\cdots, n-k-1$ span the range of $T$. You may very well need many $v_i$, but let's just say that one $v$ is enough. Then you need to prove that these elements are linearly independent (where $v$ here is an element such that $T^{n-k-1}v\neq 0$). This is not hard.

Once you do this, you are done, because you have a basis for $X$ (along with the kernel of $T$ as you remark) and all you need to do is form the matrix of $T$ with respect to the basis $\{x_1,\cdots, x_{k+1},v,\cdots T^{n-k-1}v\}$.

For instance, this matrix is (say $k=0$ and $n=3$) $$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$$ just as you wanted.

I think you mistakenly believe that the matrix you are getting is

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}$$ or something similar. This is not correct; if this were true, $T$ would carry $v$ to $v$, $Tv$ to $Tv$ etc. It does not, it shifts the basis elements. Try it yourself with $v=(0,0,1)=e_3$ and $T(e_i)=e_{i-1}$ to get a feel for the form.

In general you need to consider all $v_i$ that form so called cyclic subspaces of $X$ under $T$, take them all together with the kernel and glue the transformations together in a big direct sum. Here is a set of notes I just googled which contains details: http://www.mth.msu.edu/~shapiro/pubvit/Downloads/CycNilp/CycNilp.pdf