200 023 002/Jordan normal form/Example

We consider the matrix

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

and want to bring it to Jordan normal form. The vectors ${}u=e_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$ and ${}v=e_{2}={\begin{pmatrix}0\\1\\0\end{pmatrix}}$ are linearly independent eigenvectors to the eigenvalue ${}2$. We have

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

so that ${}u$ and ${}v$ span this eigenspace. An eigenvector must be the image of some vector under the matrix ${}A$. In fact, the linear system

${}e_{2}=Aw\,$

has the solution ${}w={\begin{pmatrix}0\\0\\{\frac {1}{3}}\end{pmatrix}}$. Therefore, the matrix ${}M$ acts in the following way

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

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

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

This matrix is in Jordan normal form with the Jordan blocks ${}{\begin{pmatrix}2\end{pmatrix}}$ and ${}{\begin{pmatrix}2&1\\0&2\end{pmatrix}}$.