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I'm having trouble finding jordan basis for matrix A, e.g. the P matrix of: [itex]J=P^{-1}AP[/itex]

Given [itex]A = \begin{pmatrix} 4 & 1 & 1 & 1 \\ -1 & 2 & -1 & -1 \\ 6 & 1 & -1 & 1 \\ -6 & -1 & 4 & 2 \end{pmatrix}[/itex]

I found Jordan form to be: [itex]J = \begin{pmatrix} -2 & & & \\ & 3 & 1 & \\ & & 3 & \\ & & & 3 \end{pmatrix}[/itex]

Now wer'e looking for [itex]v_1, v_2, v_3, v_4[/itex] such that:

[itex] Av_1 = -2v_1 → (A+2I)v_1=0[/itex]

[itex]Av_2 = 3v_2 → (A-3I)v_2=0[/itex]

[itex]Av_3 = v_2+3v_3 → (A-3I)v_3=v_2[/itex]

[itex]Av_4 = 3v_4 → (A-3I)v_4=0 [/itex]

So now I find: [itex]v_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix} \hspace{10mm} v_2,v_4 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}[/itex]

Now I try to solve [itex] (A-3I)v_3=v_2[/itex] for each of the possible v2's I just found above, but there's no solution for any of em'...

[itex]A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 1 \\ -2 \end{pmatrix}\hspace{5mm} OR \hspace{5mm} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & -1 & -1 \\ 6 & 1 & -4 & 1 \\ -6 & -1 & 4 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}[/itex]

Where am I going wrong? Thanks in advance!

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# Linear Algebra - Jordan form basis

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