# 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.

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 $$A^n = O$$.
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?