We have linear $T: V \to V$ that is $n$ dimension operator such that $T^n = 0$ and $T^{n-1} \neq 0$. Also note that $T^1 = T$ and $T^i = T^{i-1}T$. Prove that there is an ordered basis $\alpha$ of $V$ such that matrix $A = [T]_{\alpha}$ equals to the superdiagonal matrix with 1 on the superdiagonal and 0 everywhere else.
I have seen a case very similar to this where you use the direct sum but that was when $1$s were on the diagonal. Does the superdiagonal case end up with a similar proof? If you could give me a hint, I would greatly appreciate it. Thanks.
OK, time to expand on my hint.
Since $T^{n-1} \neq 0$, there exists some $v$ such that $T^{n-1} v \neq 0$. Consider the ordered list $$B = (T^{n-1} v, T^{n-2} v, \ldots, Tv, v).$$ Note that $|B| = n = \operatorname{dim} V$. In particular, $B$ is a basis if and only if $B$ is linearly independent.
Suppose $$a_0 v + a_1 Tv + \ldots + a_{n-1}T^{n-1} v = 0.$$ Apply $T^{n-1}$ to both sides. Since $T^n = 0$, we have $$a_0 T^{n-1} v = a_0 T^{n-1} v + a_1 T^n v + a_{n-1} T^{2n - 2} v = 0,$$ hence $a_0 = 0$, as $T^{n-1} v \neq 0$. Similarly, by taking $T^{n-2}$ of both sides, we can also conclude $a_1 = 0$, etc. (You can formalise this with induction.) In any case, we obtain $a_0 = a_1 = \ldots = a_{n-1} = 0$, so $B$ is linearly independent and hence a basis.
Now we compute $[T]_B$. Let $e_k$ be the $k$th standard basis column vector. We have, $$T(T^{n-1} v) = T^n v = 0 \implies [T(T^{n-1} v)]_B = \begin{bmatrix}0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}.$$ For $k > 1$, we also have $$T(T^{n-k} v) = T^{n - (k - 1)} v = 0 T^{n-1} v + 0 T^{n - 2} v + \ldots + 0 T^{n - (k - 2)} v + T^{n - (k - 1)} v + 0T^{n - k}v + \ldots + 0 v,$$ and hence $[T(T^{n - k} v)]_B = e_{k-1}$.
Putting this in a matrix, $$[T]_B = \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 \\ \end{pmatrix}.$$