How to find Jordan blocks from minimal polynomial

Question

If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?

Answer 1

From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks: $$\begin{pmatrix} \lambda & 1 & 0 & 0\\ 0 & \lambda & 1 & 0\\ 0 & 0 & \lambda & 1\\ 0 & 0 & 0 & \lambda\\ \end{pmatrix}, \begin{pmatrix} \lambda & 1\\ 0 & \lambda\\ \end{pmatrix} \mbox{or}$$ $$\begin{pmatrix} \lambda & 1 & 0 & 0\\ 0 & \lambda & 1 & 0\\ 0 & 0 & \lambda & 1\\ 0 & 0 & 0 & \lambda\\ \end{pmatrix}, \begin{pmatrix} \lambda \end{pmatrix}, \begin{pmatrix} \lambda \end{pmatrix}.$$