221 023 002/Jordan normal form/Example

We consider the matrix

${}M={\begin{pmatrix}2&2&1\\0&2&3\\0&0&2\end{pmatrix}}\,,$

and want to bring it to Jordan normal form. bringen. The vector ${}u=e_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$ is an eigenvector to the eigenvalue ${}2$. We have

${}A:=M-2E_{3}={\begin{pmatrix}0&2&1\\0&0&3\\0&0&0\end{pmatrix}}\,,$

so that there exists no further linearly independent eigenvector. We look at the linear system ${}e_{1}=Av$. This imples (looking at the second row) ${}v_{3}=0$ and so ${}2v_{2}=1$ (we can choose ${}v_{1}$ freely to be ${}0$). Hence, we set ${}v={\begin{pmatrix}0\\{\frac {1}{2}}\\0\end{pmatrix}}$. Finally, we need a solution for ${}v=Aw$. This yields the equation ${}w={\begin{pmatrix}0\\-{\frac {1}{12}}\\{\frac {1}{6}}\end{pmatrix}}$. The matrix ${}M$ acts as

$Mu=2u,\,Mv=2v+u,\,Mw=2w+v,$

so that the mapping is described with respect to the basis ${}u,v,w$ by

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

This matrix is a Jordan matrix and, in particular, in Jordan normal form.