3x3-matrix/235/Jordan/Example

We want to apply fact to

${}M={\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}\,.$

We have

${}M^{2}={\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}\cdot {\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}={\begin{pmatrix}0&0&10\\0&0&0\\0&0&0\end{pmatrix}}\,$

and

${}M^{3}=0\,.$

Therefore,

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

We have

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

so that wir can choose

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

We have

${}Me_{3}={\begin{pmatrix}3\\5\\0\end{pmatrix}}\in V_{2}\,.$

Hence,

${}V_{2}=V_{1}\oplus U_{2}\,$

with

${}U_{2}=\langle {\begin{pmatrix}3\\5\\0\end{pmatrix}}\rangle \,.$

Finally,

${}M^{2}e_{3}=M{\begin{pmatrix}3\\5\\0\end{pmatrix}}={\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}{\begin{pmatrix}3\\5\\0\end{pmatrix}}={\begin{pmatrix}10\\0\\0\end{pmatrix}}\,.$

Therefore,

${\begin{pmatrix}10\\0\\0\end{pmatrix}},\,{\begin{pmatrix}3\\5\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\0\\1\end{pmatrix}}$

is a basis with the intended properties.

The inverse matrix to

${}B={\begin{pmatrix}10&3&0\\0&5&0\\0&0&1\end{pmatrix}}\,$

is

${\begin{pmatrix}{\frac {1}{10}}&-{\frac {3}{50}}&0\\0&{\frac {1}{5}}&0\\0&0&1\end{pmatrix}},$

therefore,

${}{\begin{aligned}B^{-1}MB&={\begin{pmatrix}{\frac {1}{10}}&-{\frac {3}{50}}&0\\0&{\frac {1}{5}}&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}{\begin{pmatrix}10&3&0\\0&5&0\\0&0&1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{10}}&-{\frac {3}{50}}&0\\0&{\frac {1}{5}}&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0&10&3\\0&0&5\\0&0&0\end{pmatrix}}\\&={\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}.\end{aligned}}$