Nilpotent mapping/Sum and product/Exercise/Solution

Let

${}\varphi ={\begin{pmatrix}0&1\\0&0\end{pmatrix}}\,$

and

${}\psi ={\begin{pmatrix}0&0\\1&0\end{pmatrix}}\,.$

Because of

${}{\begin{pmatrix}0&1\\0&0\end{pmatrix}}^{2}={\begin{pmatrix}0&1\\0&0\end{pmatrix}}{\begin{pmatrix}0&1\\0&0\end{pmatrix}}={\begin{pmatrix}0&0\\0&0\end{pmatrix}}\,$

and

${}{\begin{pmatrix}0&0\\1&0\end{pmatrix}}^{2}={\begin{pmatrix}0&0\\1&0\end{pmatrix}}{\begin{pmatrix}0&0\\1&0\end{pmatrix}}={\begin{pmatrix}0&0\\0&0\end{pmatrix}}\,,$

these matrices are nilpotent. We have

${}\varphi \circ \psi ={\begin{pmatrix}0&1\\0&0\end{pmatrix}}{\begin{pmatrix}0&0\\1&0\end{pmatrix}}={\begin{pmatrix}1&0\\0&0\end{pmatrix}}\,,$

which is not nilpotent, due to wegen

${}{\begin{pmatrix}1&0\\0&0\end{pmatrix}}^{2}={\begin{pmatrix}1&0\\0&0\end{pmatrix}}\,.$

Moreover,

${}\varphi +\psi ={\begin{pmatrix}0&1\\0&0\end{pmatrix}}+{\begin{pmatrix}0&0\\1&0\end{pmatrix}}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\,$

is bijective, and, hence, not nilpotent.

To the exercise