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So far I know that in the $2 \times 2$ case, the matrices $A =$

\begin{pmatrix} 0 & a \\ 0 & 0 \\ \end{pmatrix}

\begin{pmatrix} 0 & 0 \\ a & 0 \\ \end{pmatrix}

and

\begin{pmatrix} a & a \\ -a & -a \\ \end{pmatrix}

Satisfy $A^2 = 0$, but I'm not sure if this is all of them. I don't think there should be any others.

Also, are there $2 \times 2$ matrices which satisfy $A^3 = 0$ or $A^4 = 0$ or $A^5=0$, etc... but not $A^2 = 0$? I want to say "no" but I'm having a hard time explaining why.

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2 Answers 2

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The trace and the determinant must be zero. So force the trace to be zero by taking $$A=\pmatrix{a&b\\c&-a}.$$ The square of this is zero iff $-a^2+bc=0$ so we can take $c=-a^2/b$ if $a\ne0$. (You've already done the $a=0$ case.)

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It is enough to consider matrices such that $A^2=0$.

Indeed, if $A^m v=0$ but $A^{m-1}\ne 0$, then $v, Av, A^2v,\dots , A^{m-1}v$ are linearly independent (see an easy proof here). Therefore, if $A$ a nilpotent $2 \times 2$ matrix, then $A^2=0$.

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