3x3-matrix/037/Jordan/Example

We want to apply fact to

$M=\left(\begin{array}{ccc}0& 0& 3\\ 0& 0& 7\\ 0& 0& 0\end{array}\right).$

We have

$M^2=0.$

Therefore,

$V_1=\mathrm{kern}M=⟨e_1,e_2⟩\text{ and }{V}_2=K^3.$

We have

$V_2=V_1\oplus ⟨e_3⟩,$

so that we can choose

$U_2=⟨e_3⟩.$

We have

$M{e}_3=\left(\begin{array}{c}3\\ 7\\ 0\end{array}\right)\in V_1.$

Therefore,

$V_1=⟨\left(\begin{array}{c}3\\ 7\\ 0\end{array}\right),e_1⟩=U_1.$

Hence,

$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}3\\ 7\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$

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

$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right).$