0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\dim V_1(0) = 5$.

$\endgroup$
  • $\begingroup$ In iii) you can only have $3,3,2,2$ since the eigenspace is of dimension $4$. $\endgroup$ – Christoph Apr 9 '18 at 11:12
  • $\begingroup$ Oh, I just realised that I was still taking $n=9$. That was why I got confused. $\endgroup$ – ʎpoqou Apr 9 '18 at 17:55
  • $\begingroup$ Please see math.stackexchange.com/help/merging-accounts $\endgroup$ – quid Apr 9 '18 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.