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Need help, how to find Jordan base for matrix: $A=\begin{pmatrix} -1&-1 &-2 &4 \\ 1&-3 &1 &-2 \\ 0&0&2&-8\\ 0&0 & 2&-6 \end{pmatrix}$

I found the Minimal polynomial: $(x+2)^3$

and normal Jordan is $Aj=\begin{pmatrix} -2&0 &0 &0 \\ 0&-2 &1 &0 \\ 0&0&-2&1\\ 0&0 & 0&-2 \end{pmatrix}$

Then I have problem with find $v_1, v_2, v_3, v_4 $that: $(A+2I)v_1=0 $
$ (A+2I)v_2=0 $
$ (A+2I)v_3=v_2$
$ (A+2I)v_4=v_3$
when i'm trying to find it take $v_1=(1,1,0,0) $ $ v_2=(0,0,2,1)$ and later i get contradiction

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  • $\begingroup$ If the minimal polynomial is $(x+2)^4$ than the jordan normal form is $$\begin{pmatrix} -2 & 1 & 0 & 0 \\ 0 & -2 & 1 & 0 \\ 0 & 0 & -2 & 1\\ 0& 0 & 0& -2\\ \end{pmatrix}$$ $\endgroup$ – Dominic Michaelis Mar 10 '13 at 16:08
  • $\begingroup$ no, here is one block 3x3 and one 1x1 $\endgroup$ – aiki93 Mar 10 '13 at 16:28
  • $\begingroup$ not if your minimal polynomial is $(x+2)^4$ $\endgroup$ – Dominic Michaelis Mar 10 '13 at 16:29
  • $\begingroup$ your minimal polynimal is $(x+2)^3$ $\endgroup$ – Dominic Michaelis Mar 10 '13 at 16:33
  • $\begingroup$ @DominicMichaelis, in my calculations it is $(x + 2)^3$. $\endgroup$ – Andreas Caranti Mar 10 '13 at 16:34
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Ok i have as jordan basis $$\frac{1}{3} \cdot \begin{pmatrix} 2 \\ 2\\ -4 \\ -2 \end{pmatrix}; \qquad \begin{pmatrix} 6 \\ 6\\0 \\ 0\\ \end{pmatrix}; \qquad \begin{pmatrix} 4 \\ -2 \\ -8 \\ -4 \end{pmatrix}; \qquad \begin{pmatrix} 0 \\ 0\\ 0\\ 1 \end{pmatrix} $$ Got them with that ugly algorithm.

At first we compute $(A-\lambda I)^k$ for $k=1,2,3$ which is $(A+2 I)^k$ For $k=1$ this is not so difficult we have $$\begin{pmatrix} 1 & -1 &-2&4\\ 1&-1&1&-2\\ 0&0&4&-8 \\ 0 & 0 & 2 & -4 \end{pmatrix}$$ As next we calculate $(A+2 I)^2$ $$\begin{pmatrix} 0 & 0& -3 & 6 \\ 0 & 0 & -3 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix}$$

We see that $$\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$$ is not in the kernel of $(A+2I)^2$ (this is great!). So one jordan chain is $$\begin{pmatrix} 1 & -1 &-2&4\\ 1&-1&1&-2\\ 0&0&4&-8 \\ 0 & 0 & 2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 0 \\1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\-8 \\ -4 \end{pmatrix}$$ and $$\begin{pmatrix} 0 & 0& -3 & 6 \\ 0 & 0 & -3 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 0 \\1 \end{pmatrix}= \begin{pmatrix} 6 \\ 6 \\ 0 \\ 0\\ \end{pmatrix}$$ As we only got one other eigenvector $$\begin{pmatrix} 0 \\ 0 \\2 \\ 1\\ \end{pmatrix}$$ and the minimalpolynom is $(A+ \lambda I)^3$ this is the only chain we have to calculate (what a luck).

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  • $\begingroup$ yeah, the basis is good but how to compute it? $\endgroup$ – aiki93 Mar 10 '13 at 16:41
  • $\begingroup$ the algorihm is very long and complicated, didn't you learn it ? the idea is at first computing jordan chains and then shorten them $\endgroup$ – Dominic Michaelis Mar 10 '13 at 16:47
  • $\begingroup$ @aiki93 made the chains you can shorten them now $\endgroup$ – Dominic Michaelis Mar 10 '13 at 17:11
  • $\begingroup$ Ok thanks, but i didnt learn it already $\endgroup$ – aiki93 Mar 10 '13 at 18:46

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