How to prove that a nilpotent operator has a basis representation that is strictly upper triangular? I wish to prove the following:  if $T:X\to X$ is a nilpotent (and the
    linear space $X$ is finite-dimensional), then there exists a basis
    of $X$ such that the matrix representation of $T$ is upper
    triangular with zero diagonal elements.
I was trying to prove $\{x_1,...,x_{k+1},v, Tv, ..., T^{n-k-2}v\}$ is a basis for $X$ where $x_i$ are basis for $N(T)$ and $v,..., T^{n-k-2}v$ are basis for $R(T)$ where $T^{n-k}=0$ and $n-k$ is minimal. However the matrix representation w.r.t this basis is diagonal. Is there an easy way to prove this ?
I know that for all $T:X\rightarrow X$, there exists a basis that the matrix representation is  upper triangular. How is this useful?
 A: 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
A: If an upper triangular matrix is nilpotent then its diagonal is zero and hence it is strictly upper triangular. Indeed, if $A$ and $B$ are upper triangular then $C := A\cdot B$ is also upper triangular and the diagonal of $C$ is the component-wise product of the diagonals of $A$ and $B$. 
A: The existence of a basis of which $T$ is strictly upper-triangular is equivalent to the existence of a flag $X = X_0 > X_1 > \cdots > X_n = 0$ with $T(X_i) \subset X_{i+1}$. To show the latter, note that any subspace $V$ with $T(V)\subset V$ must have $\ker T\vert V \not =0$ and thus $\text{im}\; T < V$.
