# Quick Question on Zero Matrix [closed]

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – David, colormegone, Gabriel Romon, zz20s, anomaly
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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". – user247327 Mar 18 '16 at 12:07
• Calculate the entries of $B^2$ in terms of $\alpha,\beta, \gamma,\delta$ and set each result equal to zero – Omnomnomnom Mar 18 '16 at 12:12
• Alternatively, consider what the Jordan form of $B$ can be, and find all matrices similar to those Jordan forms. – Henning Makholm Mar 18 '16 at 12:15

$B^2=0$ 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 $S$.