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Find all $2\times2$ matrices

$$B =\begin{pmatrix} \alpha & \beta\\ \gamma & \delta\\ \end{pmatrix}$$ such that $B^2$ is the zero matrix.

Any help would be greatly appreciated. Thanks

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closed as off-topic by David, colormegone, Gabriel Romon, zz20s, anomaly Mar 18 '16 at 16:06

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    $\begingroup$ Did you try? What do you get when you square that given matrix? Set each component equal to 0 and try to solve for the components of the original matrix. You will need to consider several different "cases". $\endgroup$ – user247327 Mar 18 '16 at 12:07
  • $\begingroup$ Calculate the entries of $B^2$ in terms of $\alpha,\beta, \gamma,\delta$ and set each result equal to zero $\endgroup$ – Omnomnomnom Mar 18 '16 at 12:12
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    $\begingroup$ Alternatively, consider what the Jordan form of $B$ can be, and find all matrices similar to those Jordan forms. $\endgroup$ – Henning Makholm Mar 18 '16 at 12:15
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$B^2=0$ means $\operatorname{im}(B)\subseteq\ker(B)$. In particular, $B$ needs to be of rank at most one: If $\dim\operatorname{im}(B)=2$, then $\dim\ker(B)=0$ and the above inclusion is impossible. Furthermore, $B^2=0$ is equivalent to $(SBS^{-1})^2=0$, so we can change bases as we please. Let us assume $B\ne0$ and $B^2=0$, then the kernel and image of $B$ are both of dimension one. We change bases so that $\ker(B)$ is spanned by $$e_1=\begin{pmatrix}1\\0\end{pmatrix}.$$ Then, $$\exists a,b\colon\quad B=\begin{pmatrix} 0 & a \\ 0 & b \end{pmatrix}.$$ Since $\operatorname{im}(B)\subseteq\ker(B)$, we have $b=0$. As $B\ne0$, we have $a\ne 0$. By a change of bases, we can achieve that $a=1$, because $$ \begin{pmatrix} a^{-1} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$

Hence, aside from the zero matrix, every matrix $B$ with $B^2=0$ is of the form $$S \cdot \begin{pmatrix} 0 & 1 \\ 0 &0\end{pmatrix}\cdot S^{-1}$$ for an invertible matrix $S$.

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