# Finding Jordan form when characteristic is degree 8 and minimal is degree 4.

I'm hoping someone can help me with this question and see if I understand the theory behind it.

Suppose we have a linear transformation $T$ represented by a matrix $A$. Let the characteristic polynomial $C_A(x) = (\lambda - 1)^8$ and the minimal polynomial $\mu_A(x) = (\lambda -1)^4$. What are the possible JCF's of A? How would you dedide which is the correct one?

My response / understanding: So since the minimal polynomial is of degree 4, we have guaranteed a 4x4 block which we can put in the top left hand corner. The possible JCF forms of the other 4x4 block is dependent on the dimension of the eigenspace so it could be any combination, such as (but not only) \begin{bmatrix} 1 & 1 & 0 & 0 \\0 & 1 &0 &0 \\ 0 & 0 & 1 & 0 \\ 0& 0 &0 &1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 & 0 \\0 & 1 &1 &0 \\ 0 & 0 & 1 & 0 \\ 0& 0 &0 &1 \end{bmatrix}

The second part I am having more difficulty with. My strategy is to first find the prejordan basis, then convert this into a Jordan basis and look at my individual Jordan chains. The length of these Jordan chains should indicate the size of each jordan block. Whilst I think this method works, I would appreciate a more theoretical explanation on its effectivness.

Thank you

• To get the JCF it is enough to calculate the dimensions of $\ker(A-I)^n$ for $n=1,2,3$ (at most). – A.Γ. Dec 30 '15 at 16:05

For $d\geq 1$ let $J_\lambda^{(d)}$ be the $d\times d$ matrix $$J_\lambda^{(d)}= \begin{bmatrix} \lambda & 1 & \\ & \lambda & 1 & \\ && \lambda & 1 & \\ &&& \ddots & \ddots & \\ &&&& \lambda & 1 \\ &&&&& \lambda & 1 \\ &&&&&& \lambda \end{bmatrix}$$ where all the unmarked entries are $0$.

For square matrices $A_1,\dotsc,A_n$ let $A_1\oplus\dotsb\oplus A_n$ be the block-diagonal matrix $$A_1\oplus\dotsb\oplus A_n = \begin{bmatrix} A_1 \\ & A_2 \\ &&\ddots \\ &&& A_n \end{bmatrix}$$ where again the unmarked entries are $0$.

Now, we are given that $A$ has characteristic polynomial and minimal polynomial \begin{align*} \chi_A(t) &= (t-1)^8 & \mu_A(t) &= (t-1)^4 \end{align*} respectively. This tells us that the Jordan form of $A$ is of the form $$J=J_1^{(d_1)}\oplus \dotsb\oplus J_1^{(d_s)}$$ where

• $d_1\geq\dotsb\geq d_s$
• $d_1+\dotsb+d_s=\deg\chi_A(t)=8$
• $\max\{d_1,\dotsc,d_s\}=d_1=\deg\mu_A(t)=4$

Thus the number of possible Jordan forms is the number of positive integer solutions to $$d_2+\dotsb+d_s=4$$ All the possible solutions are given by \begin{align*} 4 &= 4 & 3+1 &=4 & 2+2 &= 4 & 2+1+1 &= 4 & 1+1+1+1 &= 4 \end{align*} Thus all possible Jordan forms are \begin{align*} J_1^{(4)}\oplus J_1^{(4)} && J_1^{(4)}\oplus J_1^{(3)}\oplus J_1^{(1)} && J_1^{(4)}\oplus J_1^{(2)}\oplus J_1^{(2)} \\ J_1^{(4)}\oplus J_1^{(2)}\oplus J_1^{(1)}\oplus J_1^{(1)} && J_1^{(4)}\oplus J_1^{(1)}\oplus J_1^{(1)}\oplus J_1^{(1)}\oplus J_1^{(1)} && \end{align*} Without more information we cannot determine which of these is correct.

However, we can figure out which Jordan form is correct by looking at the invariants \begin{align*} \dim\ker(A-I) && \dim\ker(A-I)^2 && \dim\ker(A-I)^3 \end{align*} Here's the exhaustive list of possibilities:

\begin{array}{c|c|c|c} J & \dim\ker(J-I) & \dim\ker(J-I)^2 & \dim\ker(J-I)^3 \\ \hline J_1^{(4)}\oplus J_1^{(4)} & 2 & 4 & 6 \\ J_1^{(4)}\oplus J_1^{(3)}\oplus J_1^{(1)} & 3&5&7 \\ J_1^{(4)}\oplus J_1^{(2)}\oplus J_1^{(2)} &3&6&7 \\ J_1^{(4)}\oplus J_1^{(2)}\oplus J_1^{(1)}\oplus J_1^{(1)}&4&6&7 \\ J_1^{(4)}\oplus J_1^{(1)}\oplus J_1^{(1)}\oplus J_1^{(1)}\oplus J_1^{(1)}&5&6&7 \end{array} For more information about the minimal polynomial and Jordan canonical form see Tom Leistner's notes on the topic.

• Thank you for the response, however I understand the first part of the question and was looking for more information on a method (and the corresponding theory) on how to determine which form is correct. – Kevin Dec 31 '15 at 1:39
• @Kevin Ok updated. – Brian Fitzpatrick Dec 31 '15 at 1:58
• Thanks Brian this is making a lot of sense. Am I correct in saying that every Jordan chain must be 'finished' in the sense that there must always be at least one vector in the dimension ker(J-I)? And this is why we couldn't have something like dim ker(J-I) = 2, dim ker(J-1I)^ = 3, dim ker(J-I)^3 = 7? – Kevin Jan 2 '16 at 16:02