Show that $A^n = 0$ but $A^{n-1} \neq 0$ for an $n\times n$ strictly lower triangular matrix How can I show that for the matrix:
$$A = \begin{bmatrix}0 & 0 &\ldots &0 &0 \\
1 & 0&\ldots &0 &0 \\
0 & 1 & \ldots & 0 &0\\
\vdots & \ddots &\ddots&\ddots&\vdots \\
0&0&\ldots&1&0\end{bmatrix} = \begin{bmatrix} \vec{e}_2&\vec{e}_3&\ldots&\vec{e}_n&\vec{0} \end{bmatrix}$$
$A^n = 0$ but $A^{n-1} \neq0$? I can see it in example cases when $n=1,2,3$ but I'm struggling to piece together what is actually happening.
 A: Programme for solution:


*

*Spell out what $A^2$ is.

*Spell out what $A^3$ is.

*...

*Continue until you guess what $A^k$ is. Q.E.D.

A: Hint:
Denoting $a_{ij}^k$ the lelemnt in row $i$, column $j$ of $A^k$, show that  each $A^k$ has coefficients equal to $1$ on its $k-1$-th subdiagonal, $0$ elsewhere:
$$\forall k>0,\,
\begin{cases}
a_{ij}^k=1&\text{if }i-j=k, \\[1ex]
a_{ij}^k=0&\text{otherwise}
\end{cases}$$
A: As a linear mapping, the $A$ actually takes $e_j$ to $e_{j+1}$ for $j\leqslant n-1$, and takes $e_n$ to $0$, so as matrix products,  for each $k\in \{1,2,\dots,n\}$, $A^j e_k = e_{k+j}$ for $j = 1, 2, \dots, n-k$ and then $A^{n-k+1}e_k =0$, hence $A^p e_k=0$ for $p \geqslant n-k+1$.  Put $e_j$'s in a row, and note that $\begin{bmatrix} e_1 & e_2 & \cdots & e_n\end{bmatrix} = I_n$ is an $n \times n$ identity matrix, we have 
$$
A = A \begin{bmatrix} e_1 & e_2 & \cdots & e_{n-1} & e_n\end{bmatrix} = \begin{bmatrix} Ae_1 & Ae_2 & \cdots & Ae_n\end{bmatrix}= \begin{bmatrix} e_2 & e_3 & \cdots & e_n & 0\end{bmatrix},
$$
and
$$
A^k =A^k \begin{bmatrix} e_1 & e_2 & \cdots & e_n\end{bmatrix} = \begin{bmatrix} A^ke_1 & A^ke_2 & \cdots & A^ke_n\end{bmatrix} =  \begin{bmatrix} e_{k+1} & e_{k+2} & \cdots & e_n & 0 &\cdots & 0\end{bmatrix}, 
$$
specifically
$$
A^{n-1} = A^{n-1}\begin{bmatrix} e_1 & e_2 & \cdots & e_n\end{bmatrix} =  \begin{bmatrix} e_n & 0& \cdots & 0\end{bmatrix} \neq 0
$$
while $A^n = A^n \begin{bmatrix} e_1 & e_2 & \cdots & e_n\end{bmatrix} = 0$. 
A: Recall that, for a matrix $A$, the $k$th column of $A$ is $Ae_k$ where $e_k$ is the standard basis (column) vector of the appropriate size, with $1$ in the $k$th position, and $0$s elsewhere.
In this way, we see that $Ae_1 = e_2$, $Ae_2 = e_3$, and more generally, $Ae_k = e_{k+1}$ for all $1 \le k < n$. Therefore, $A^{n-1} e_1 = e_n \neq 0$, so $A^{n-1} \neq 0$.
On the other hand, given the last column is $0$, we have $Ae_n = 0$, and given any $k \in \{1, \ldots, n\}$, we have
$$A^{n - k + 1} e_k = A^{n - k + 1} A^{k - 1} e_1 = A^n e_1 = Ae_n = 0.$$
Thus,
$$A^n e_k = A^{k - 1} A^{n - k + 1} e_k = A^{k - 1} 0 = 0.$$
In other words, the linear transformation $x \mapsto A^nx$ maps all the standard basis vectors to $0$, which means it must be the $0$ transformation, which can only happen if $A^n = 0$.
