find a matrix $A_{n \times n}$ which satisfy $A^{n}=0$ and $A^{n-1}\ne0$ I asking a question for my brother. He just started his algebra course not long ago and he got a question he is stuck on (he just started the course so all his knoweldge is based on matrices and their properties).
I need to find a Matrix $A_{2x2} \not=0$ which satisfies $A^2=0$.
Also, I need to find a Matrix $A_{3x3}$ which satisfies $A^3=0$ and $A^2\not=0$.
In the end I need to generalize the problem to $A_{n \times n}$ (find a matrix $A_{n \times n}$ which satisfy $A^{n}=0$ and $A^{n-1}\ne0$).
In the first question I wrote $A$ as
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
after multiplying $A$ with itself I found out that every matrix $2x2$ in the form of
\begin{bmatrix} x & y \\ -\frac{x^2}{y} & -x \end{bmatrix}
when multiplied with itself equal zero.
In the second question I tried the same thing but it got to long to soon so I figured I am doing something wrong. also, the the equations I get when I am trying to find conditions for $A$ are to messy to deal with (9 in total).
I did manage to find some matrices that satisfies $A^3=0$ such as
\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}
but I still wish to know how should I find the conditions for it to happen.
About the third question, I guess I need to finish the second one to even begin thinking about the solution but I still can't see where should I start.
 A: Try this:
$$A = \begin{bmatrix}
\color{red}0 & 1 & 1 & ... & 1 \\
0 & \color{red}0 & 1 & ... & 1 \\
... & ... & ... & ... & ...\\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 & ... & \color{red}0 \end{bmatrix} $$

Just to feel what will happen:
$$A^2 = \begin{bmatrix}
0 & 1 & 1 & ... & 1 \\
0 & 0 & 1 & ... & 1 \\
... & ... & ... & ... & ...\\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 & ... & 0 \end{bmatrix}\begin{bmatrix}
0 & 1 & 1 & ... & 1 \\
0 & 0 & 1 & ... & 1 \\
... & ... & ... & ... & ...\\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 & ... & 0 \end{bmatrix} $$$$= 
\begin{bmatrix}
\color{red}0 & \color{blue}0 & 1 & 2&  ... & n-1 \\
0 & \color{red}0 & \color{blue}0 & 1& ... & n-2 \\
... & ... & ...&... & ... & ...\\
0 & 0 & 0 & 0& ... & \color{blue}0 \\
0 & 0 & 0 & 0&... & \color{red}0 \end{bmatrix}$$
Each time you raise to another power, the upper diagonals will start to vanish consequently (actually, you can find an explicit formula for the $n^\text{th}$ power quite easily). Then, notice that $A^{n-1} \ne O$ but $A^n = O$.
A: If you think about the first part in linear transformation terms, you want a map $A: \Bbb{R}^2 \to \Bbb{R}^2$ for which $A(A \vec{v}) = \vec{0}$ for any vector $\vec{v}$.
One option for such a transformation would be $$\langle x, y \rangle \mapsto \langle 0, x \rangle.$$ What would be the matrix of this transformation?
