# Finding Jordan basis of a matrix ($3\times3$ example)

Our teacher didn't explain us how to find it so I've had to look up a bit by myself.

I have this matrix :

$$A = \begin{pmatrix} 9 & 4 & 5 \\ -4 & 0 & -3 \\ -6 & -4 & -2 \end{pmatrix}$$

I've found its characteristic polynomial $p_A(\lambda) = -(\lambda -3)(\lambda - 2)^2$.

I've found its eigenspaces : $E_2 = span(\begin{pmatrix}-2 & 1 & 2\end{pmatrix}^T)$, $E_3 = span(\begin{pmatrix}-3 & 2 & 2\end{pmatrix}^T)$.

And I've found the generalized eigenspace of $\lambda = 2$ : $\ker((A-2I_3)^2) = span(\begin{pmatrix}0 & 1 & 0\end{pmatrix}^T, \begin{pmatrix}1 & 0 & -1\end{pmatrix}^T)$

But I don't know what to do now. I've tried to see what result I would get with $$M = \begin{pmatrix} -3 & -2 & 0 \\ 2 & 1 & 1 \\ 2 & 2 & 0 \end{pmatrix}.$$

But $$M^{-1}AM = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 2 & -2 \\ 0 & 0 & 2 \end{pmatrix}.$$

So there it is. Any help ?

## 1 Answer

What you did is very close to correct.

For the generalized eigenvector, call it $v$, you want not just something that is in $\ker((A-2I)^2)$, but also you want $$(A-2I)v=\begin{pmatrix} -2\\ 1\\ 2\\ \end{pmatrix}$$

This is what causes the 1 to appear above the diagonal.

If you modify your vector slightly, you get the desired result.

Let $$v = \begin{pmatrix} 0\\ -\frac{1}{2}\\ 0 \end{pmatrix}_.$$

Then things will work.