Nilpotent mapping/Sum and product/Exercise/Solution

Let

$\phi =\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$

and

$\psi =\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right).$

Because of

${\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)}^2=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$

and

${\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)}^2=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),$

these matrices are nilpotent. We have

$\phi \circ \psi =\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),$

which is not nilpotent, due to wegen

${\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)}^2=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right).$

Moreover,

$\phi +\psi =\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)+\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

is bijective, and, hence, not nilpotent.

To the exercise