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I have to find the Jordan basis of the following matrix:

$$A=\begin{pmatrix} 2&0&-1&1\\ 1&1&-1&1\\ 1&-3&2&-1\\ 0&-3&2&-1\\ \end{pmatrix} $$

I computed the characteristic polinomial $P_A(t)=(t-1)^4$

I'll just skip the calculations and state what I found, I got (${\lambda_1=1}$): $$\dim(V_{\lambda_1})=\dim(\ker(A-\lambda_1I))=2$$$$V_{\lambda_1}^2=\ker((A-\lambda_1I)^2)=Span\left(\begin{pmatrix} 0\\ 1\\ 0\\ 0\\ \end{pmatrix}\begin{pmatrix} 1\\ 0\\ 1\\ 0\\ \end{pmatrix}\begin{pmatrix} -1\\ 0\\ 0\\ 1\\ \end{pmatrix}\right)$$$$V_{\lambda_1}^3=\mathbb{R^4}$$ Therefore for my Jordan basis: $$v_3\in V_{\lambda_1}^3/V_{\lambda_1}^2$$ So I choose $v_3=\begin{pmatrix} 0\\ 0\\ 0\\ 1\\ \end{pmatrix} $

Then I calculated $v_2$ as $v_2=(A-\lambda_1I)v_3$ and $v_1$ as $v_1=(A-\lambda_1I)^2v_3$.

I got $$v_2=\begin{pmatrix} 1\\ 1\\ -1\\ -2\\ \end{pmatrix},\;\;v_1=\begin{pmatrix} 0\\ 0\\ -1\\ -1\\ \end{pmatrix}$$ Indeed these 3 vectors are fine. (I checked by computing) When I have to choose the last vector $v_4$ (i.e. the one associated with the Jordan block of order 1) I'm not sure how to procede, I read on some notes that I should choose $v_4 \in V_{\lambda_1}^3/span(v_1,v_2,v_3)$ but that doesn't really seam to work...

The method presented in my textbook is somewhat unclear and I couldn't find anywhere a good explaination on a canonical method to follow when looking for the Jordan Basis of a matrix...

Spasoje Durovic
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  • The advantage of this site is that your question has appeared here already many times, and people have given great answers and help for this. Have a look. Start for example with this post. – Dietrich Burde Jan 08 '20 at 16:13
  • Does this answer your question? Finding Jordan basis of a matrix (4x4) – Alejandro Tolcachier Jan 08 '20 at 16:21
  • @AleTolcachier Yeah, I'm carefully reading the pdf linked in the comment section of that post, still it would've been nice to recive a direct answer since I should be pretty close to the conclusion, but... it's fine :) – Spasoje Durovic Jan 08 '20 at 16:24
  • Quick update, I just read a pdf linked in the comments of the other question about JCF, basically in my attempt I had selected a generalized eighenvector of rank 3, one of rank 2 and one of rank 1, since our eighenvalue have algebraic multiplicity 4 we have exactly 4 linearly independent generalized eigenvectors, noticing that, it must be the case that the last generalized eighenvector has rank 1 since the rank 1 eighenspace has dimension two (and thus contain two linearly independent eighenvectors), so for $v_4$ we can select $(1,2/3,1,0)^T$ which is l.i. from the other selected vectors – Spasoje Durovic Jan 08 '20 at 17:53

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