3x3-matrix/037/Jordan/Example

We want to apply fact to

${}M={\begin{pmatrix}0&0&3\\0&0&7\\0&0&0\end{pmatrix}}\,.$

We have

${}M^{2}=0\,.$

Therefore,

$V_{1}=\operatorname {kern} M=\langle e_{1},e_{2}\rangle {\text{ and }}V_{2}=K^{3}.$

We have

${}V_{2}=V_{1}\oplus \langle e_{3}\rangle \,,$

so that we can choose

${}U_{2}=\langle e_{3}\rangle \,.$

We have

${}Me_{3}={\begin{pmatrix}3\\7\\0\end{pmatrix}}\in V_{1}\,.$

Therefore,

${}V_{1}=\langle {\begin{pmatrix}3\\7\\0\end{pmatrix}},e_{1}\rangle =U_{1}\,.$

Hence,

${\begin{pmatrix}1\\0\\0\end{pmatrix}},\,{\begin{pmatrix}3\\7\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\0\\1\end{pmatrix}}$

is a basis as looked for. With respect to this basis, the linear mapping is described by the matrix

${\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}}.$