Let V be an n-dimensional vector space, and T from V to V a linear transformation... Let $V$ be an $n$-dimensional vector space, and $T$ from $V$ to $V$ a linear transformation. Show that $T$ is nilpotent of order $n$ if and only if there exists a basis $\beta = {v_1,v_2,....v_n}$ of $V$ such that the matrix of $T$ relative to $\beta$ is of the form $$\begin{pmatrix} 0 & 1 & 0... & 0 \\ 0 & 0 & 1... & 0 \\ \vdots& \vdots & \vdots & \vdots\\ 0& 0 & 0... & 1\\ 0 & 0 & 0... & 0 \end{pmatrix}$$ (Sorry i can't figure out how to show the "..." vertically!)
I know that because it says "if and only if" i have to show both ways. I started assuming that $T$ is nilpotent, knowing this means the eigenvalues are zero, and trying to show there existed a basis of V such that the matrix of $T$ relative to $\beta$ is of that form, but i'm not sure how to show that the basis would be relative to $\beta$? I think i'm a little confused on how to approach this.
 A: If $T$ is a nilpotent of order $n$, then there exists a nonzero vector 
$v_n\in V$ such that $T^n(v_n)={\it 0}$ and $T^i(v_n)\ne{\it 0}$
for all $1\leq i\leq n-1$. So we consider the set 
$$\beta=\left\{T^{n-1}(v_n),T^{n-2}(v_n),\ldots,T(v_n),v_n\right\}$$
and we show that $\beta$ is the desired ordered basis for $V$. First,
we claim that $\beta$ is linearly independent. Given the scalars 
$a_1,a_2,\ldots,a_n$ such that
$$\sum_{i=1}^na_iT^{n-i}(v_n)={\it 0}.$$
Then
$${\it 0}
=T^{n-1}({\it 0})
=T^{n-1}\left(\sum_{i=1}^na_iT^{n-i}(v_n)\right)
=\sum_{i=1}^na_iT^{2n-i-1}(v_n)
=a_nT^{n-1}(v_n),$$
it follows that $a_n=0$ and original equation becomes
$$\sum_{i=1}^{n-1}a_iT^{n-i}(v_n)={\it 0}.$$
Then we use the same way to observe
$${\it 0}
=T^{n-2}({\it 0})
=T^{n-2}\left(\sum_{i=1}^{n-1}a_iT^{n-i}(v_n)\right)
=\sum_{i=1}^{n-1}a_iT^{2n-i-2}(v_n)
=a_{n-1}T^{n-1}(v_n),$$
which follows that $a_{n-1}=0$. Continue the process we may deduce that $a_i=0$ for each $i$, and hence $\beta$ is linearly independent, as
claimed. Since $V$
is $n$-dimensional, $\beta$ is actually an ordered basis for $V$. Finally, denote $v_i=T^{n-i}(v_n)$ for $1\leq i\leq n-1$,
it is clear that $T(v_{i+1})=v_i$ for all $1\leq i\leq n-1$, and thus
$$[T]_\beta=
\begin{pmatrix}
0&1&0&\cdots&0&0\\
0&0&1&\cdots&0&0\\
0&0&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&0&1\\
0&0&0&\cdots&0&0
\end{pmatrix}.$$
The converse is immediate.
