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I have a $6 \times 6$ matrix with characteristic polynomial $(x-1)^6$ and minimal polynomial $(x-1)^3$, I'm asked to find basis matrix P consisting of generalized eigenvectors such that $P^{-1}AP=J$. I started with a vector $v_3 \in Ker(A-I)^3 \setminus ker(A-I)^2$. With this I'm able to get three basis vectors say $v_1,v_2,v_3$ where: $v_2=(A-I)v_3$ and $v_1=(A-I)v_2$. How to get others? Any help is highly appreciated. Note: GM of Eigen value is 4.

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  • $\begingroup$ What dimensions do you get for $\ker(A-I)^k$ for $k=1,2,3$? $\endgroup$
    – Berci
    Commented Oct 6, 2021 at 11:28
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    $\begingroup$ dim(Ker(A-I))=GM of $\lambda=1 =4$. Others are $5,6$ for $K=2,3$ respectively. $\endgroup$
    – UserA
    Commented Oct 6, 2021 at 11:46

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With the given data of characteristic and minimal polynomials, we must understand that the Jordan Form is not unique. There are 3 possible Forms corresponding to 3+3,3+2+1 and 3+1+1+1 cases. For the 3+3 case the Jordan form is $$ \begin{pmatrix} 1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 0&0&1&0&0&0&\\ 0&0&0&1&1&0\\ 0&0&0&0&1&1\\ 0&0&0&0&0&1\\ \end{pmatrix} $$ and hence a basis is {$e_1,e_1+e_2,e_2+e_3,e_4,e_4+e_5,e_5+e_6$}. We have to work out for the other two cases similarly.

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  • $\begingroup$ I appreciate your reply. But sir I have mentioned in note that geometric multiplicity is 4. Which leaves only one choice for JC form upto rearrangement of blocks, i,e $3+1+1+1$.moreover I have not asked to find basis of J but basis matrix P such that $P^{-1}AP=J$. $\endgroup$
    – UserA
    Commented Oct 6, 2021 at 11:56
  • $\begingroup$ GM=4 you have given as note. Didn’t see that important point in the problem. $\endgroup$ Commented Oct 6, 2021 at 22:55

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