Question

Show that $A=\left[\begin{array}{cc}ab& b^2\\ {-a}^2& -ab\end{array}\right]$ is nilpotent of index $2$.
Show as above that
$A^2=A.A=\left[\begin{array}{cc}{a}^2{b}^2-& a^2{b}^2\\ {-a}^{3}b+& a^{3}b\end{array}\begin{array}{cc}a{b}^3-& {ab}^3\\ {-a}^2{b}^2+& a^2{b}^2\end{array}\right]$ $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=O$
Hence, $A$ is nilpotent of index $2$.

Answer 1

$A=\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right]A^2=\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right]\left[\begin{array}{cc}ab& b^2\\ -a^2& -ab\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$A^2=0$

$\therefore A$ is nitpotent with index $2$