# Possible Jordan Normal Forms

Let $T : \Bbb C_n \to \Bbb C_n$ be a linear map and suppose that $ch_T (x) = x^n$. What are the possible Jordan normal forms of T given the information:

(i) $n = 6 \text{ and } m_T (x) = x^4.$

(ii) $n = 9,\ m_T (x) = x^5 \text{ and } dim(V_1(0)) = 3.$

(iii) $n = 10,\ m_T (x) = x^3,\ dim(V_1(0)) = 4 \text{ and } dim(V_2(0)) = 8.$

Here $V_i(\lambda_j)$ denotes the $i^{th}$ generalised eigenspace of the eigenvalue $\lambda_j$

What I have so far is for

i) I get matrices with the Jordan Blocks $J_4,J_1, J_1$ (3 possible arrangements with these) and $J_4, J_2$ (2 possible arrangements here). I understand that there will be at least one 4x4 Jordan block present in the Jordan Normal form, is that right?

ii) More combinations here but by the same idea. $J_5,J_2,J_2$ (3 combos). $J_5,J_3,J_1$(6 combos). Here I think there needs to be at least one 5x5 block but the eigenvalue $0$ will have 3 contributions which means exactly 3 blocks can come from it, correct?

iii) Again, $J_3,J_3,J_3,J_1$ and this gives 4 combos but I am not sure about this one. What is the significance of knowing the dimension of the $2^{nd}$ generalised eigenspace?

## 1 Answer

i) Yes, that's right. The minimal polynomial imposes the existence of a $4 \times 4$ Jordan block.

ii) Ok.

iii) No. In fact, $\dim V_2(0)=8$ means that we should have 4 blocks of size $\geq 2$, containing one vector $v$ such that $T(v)\neq 0$ but $T^2(v)=0$. Moreover, as before, $m_T$ imposes at least one $J_3$. Thus we need at least $J_3$ and 3 $J_k$ with $k\geq 2$. I can see only $3,2,2,2,1$ or $3,3,2,2$.

Edit: in fact, Christoph is right in his comment below: $3,2,2,2,1$ is impossible since it would imply $\dim V_1(0) = 5$.

• In iii) you can only have $3,3,2,2$ since the eigenspace is of dimension $4$. – Christoph Apr 9 '18 at 11:12
• Oh, I just realised that I was still taking $n=9$. That was why I got confused. – ʎpoqou Apr 9 '18 at 17:55
• – quid Apr 9 '18 at 21:55