constructing a matrix such its square is not '0' but its cube is.

Question

i have been asked to construct a matrix A such that $A^2$ is not equal to '0' but, $A^3=0$. how should i proceed.

i can only understand that all the eigenvalues for A , $A^2$ and $A^3$ will be 'zero' but then how to proceed !

please don't give me the answer only. i want to know how to proceed !

I TRIED CAYLEY HAMILTON THEOREM . (if $\lambda $ is the eigenvalue of A)

if $A^3=0 =>(\lambda)^3 =0 => (\lambda)=0 =>(\lambda)^2=0 =>A^2=0 $ so this is not possible . is there any fallacy in my application cayley hamilton. please point me out.

Answer 1

First the minimal polynomial of the matrix is $x^3$, hence the matrix has dimension $\ge3$.

With the Jordan canonical form, you can have this: $$A=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix},\quad A^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},\quad A^3=0.$$ Any matrix that is similar to $A$ will have the same property.