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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.

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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 $A^n = O$.

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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?

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  • $\begingroup$ Thank you for the help. As I have mentioned before I just started the course so all my knoweldge is based on matrices and their properties. $\endgroup$ Dec 5 '20 at 19:41

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