# Finding basis consisting of generalized Eigen vectors in Jordan normal form.

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.

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

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.
• 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 $P^{-1}AP=J$. Commented Oct 6, 2021 at 11:56