Let $V$ be an $n$-dimensional $\mathbb{k}$-vector space. Let $T$ an element of $End_{\mathbb{k}}(V)$ be an endomorphism, and $T^k := T \circ \dots \circ T$ ,($k$ times) and set $T$ to the power of $0$ as $id_V$ , where $id_V$ is the identity operator on $V$.
Now, suppose that $T^n = 0$ but $T^{n-1} \neq 0$ and let $v_0$ an element of $V$ be a fixed vector such that $T^i(v_0) \neq 0$ for $0\leq i \leq n-1$.
Show that $\{v_0, T(v_0), \dots T(v_0)^{n-1}\}$ forms an ordered basis of $V$ . Give also the coordinate matrix $A(T)$ which is an element of $Mat(n,\mathbb{k})$ of $T$ relative to this basis.
Since $\mathcal{B} = \{T^i(v_0)\}_{0 \leq i \leq n-1}$ has $n$ elements, it suffices to show linear independence to prove that it is a basis of $V$. Suppose that we have a linear combination
$$ \sum_{i=0}^n\alpha_iT^i(v_0) = 0 $$
with $\alpha_0 \dots \alpha_{n-1} \in \mathbb{k}$. Now, applying $T^{n-1}$ to both sides,
$$ 0 = T^{n-1}(0) = \sum_{i=0}^n\alpha_iT^{(n-1 + i)}(v_0) = \alpha_0T^{n-1}(v_0) $$
since every other term is a power of $T$ greater than $n$, and we're assuming $T^k = 0$ if $k \geq n$. Since $T^{n-1}(v_0) \neq 0$, we have $\alpha_0 = 0$. Hence,
$$ \sum_{i=1}^n\alpha_iT^{(n-1 + i)}(v_0) = 0 $$
Applying $T^{n-2}$ to both sides, we cancel out every term but $\alpha_1T^{n-1}(v_0)$, which again implies $\alpha_1 = 0$. With this process, at each set we arrive at $\alpha_i = 0$, which allows us to derive $\alpha_{i+1} = 0$ by now composing with $T^{n-1-(i+1)}$. Therefore $\alpha_0 = \dots \alpha_{n-1} = 0$ which proves that $\mathcal{B}$ is linearly independent, and in particular, a basis for $V$. Regarding the matrix of $T$ in this basis, if we note $w_i = T^{i}(v_0)$, it is immediate that $T(w_i) = w_{i+1}$ if $i \leq n-1$ and $T(w_{n-1}) = 0$. Thus, the matrix $[T]_{\mathcal{B}}$ will have zeroes in every entry expect the ones below the diagonal,
$$ ([T]_{\mathcal{B}})_{ij} = \delta_{i,(j+1)} $$