# Repeated Eigenvalues

Two problems from Differential Equations; Dynamical Systems, and an Introduction to Chaos (Morris W. Hirsch,Stephen Smale.Robert L. Devaney), examples page 112,113:

If $$A= \begin{pmatrix} 2 & 0 & -1 \\ 0 & 2 & 1 \\ -1 & -1 & 2 \\ \end{pmatrix}$$ … $\lambda =2 , m_{\lambda}=3$ $rank(A-2I)=2 , n_{\lambda}=1 ,k= m_{\lambda}- n_{\lambda}+1=3$ so $$p= \begin{pmatrix} 1& 0 & 1 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \\ \end{pmatrix}$$ by $v_2=(A-2I)v_3, v_1=(A-2I)v_2, v_3 : \begin{cases} (A-2I)^3v_3= 0\\ (A-2I)^2v_3 \not = 0\\ \end{cases}$But $p^{-1}Ap \not =J$ $$p= \begin{pmatrix} 0& 0 & 0 \\ 0& 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \not = \begin{pmatrix} 2& 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \\ \end{pmatrix}$$ what’s wrong?!

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I don't see a jordan algebra anywhere. Are the tags correct? – Jyrki Lahtonen Jun 27 '13 at 18:23
@JyrkiLahtonen matrix $J$ is jordan form of $A$ – agustin Jun 27 '13 at 18:25
Jordan algebra means something else. See the link. For questions about Jordan forms the recommended tag is matrices. – Jyrki Lahtonen Jun 27 '13 at 18:31

I will assume you understand how they derived $P$. (If not, please respond).

We have:

$$P= \begin{pmatrix}1& 0 & 1 \\-1 & 0 & 0 \\0 & -1 & 0 \end{pmatrix}$$

This give us:

$$P^{-1}= \begin{pmatrix} 0 & -1 & 0 \\0 & 0 & -1 \\1 & 1 & 0 \end{pmatrix}$$

From this we get:

$$P^{-1}AP = \begin{pmatrix} 0 & -1 & 0 \\0 & 0 & -1 \\1 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix}2 & 0 & -1 \\0 & 2 & 1 \\-1 & -1 & 2 \end{pmatrix} \cdot \begin{pmatrix}1& 0 & 1 \\-1 & 0 & 0 \\0 & -1 & 0 \end{pmatrix} = \begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\0 & 0 & 2\end{pmatrix}$$

This validates the authors' solution.

Did you calculate $P^{-1}$ or the product $P^{-1}AP$ incorrectly?

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ok that's right. $p^{-1}$ is wrong by me. – agustin Jun 28 '13 at 6:53
Day was slowish...but okay. At least we can say TGIF!! – amWhy Jun 29 '13 at 1:03
Yes...dead-pan slow (and likely slow weekend ahead of us!) – amWhy Jun 29 '13 at 1:06