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Given that the jordan form of the matrix $A \in Mat_7(\mathbb F)$ is:

$\begin{pmatrix} J_2(1) &\cdots &0\\0& \cdots J_3(1) \cdots &0\\0&0& \cdots J_2(2)\end{pmatrix}$

Find the Jordan form of $A^2+A+I$

Clarification: $J_{\alpha}(\beta)$ is a jordan block of size $\alpha$ with $\beta$ on the diagonal.

What I did:

I tried bruteforcing it, as in actually calculating $A^2+A+I$, but now to find the minimal polynomial of a 7x7 matrix...not fun. Is there a smarter way to go about this question?

From the jordan form of $A$ I can infer the minimal polynomial and characteristic polynomial of $A$. Does that have any implication on the characteristic and minimal polynomials of $A^2+A+I$?

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  • $\begingroup$ A tiny hint that may or may not be of specific use to you: $(A^2+A+I)(A-I)=A^3-I$... $\endgroup$ – Steven Stadnicki Feb 4 '14 at 19:28
  • $\begingroup$ I wonder if that's applicable here. $\endgroup$ – Oria Gruber Feb 4 '14 at 19:39
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Hint: If $$ A = SJS^{-1} $$ then

$$A^2+A+I = \left(SJS^{-1}\right)^2+\left(SJS^{-1}\right)+SS^{-1} = S\left(J^2+J+I\right)S^{-1}$$

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  • $\begingroup$ Brilliant!!!!...But that doesn't help that much. Now I need to find the Jordan form of $J^2+J+I$, thats still a 7 by 7 matrix $\endgroup$ – Oria Gruber Feb 4 '14 at 19:29
  • $\begingroup$ $(J^2+J+I)$ has the same amount of "blocks" as $J$ and you only have to handle each of those blocks independently. Those blocks are either Jordan blocks, or triangular and easy to get into JNF $\endgroup$ – user127.0.0.1 Feb 4 '14 at 19:47
  • $\begingroup$ So I don't need to actually find the Jordan form of $J^2+J+I$? I can find the jordan form of each block individually and infer from that the jordan form of the matrix im looking for? $\endgroup$ – Oria Gruber Feb 4 '14 at 19:50
  • $\begingroup$ Yes, that's right. $\endgroup$ – user127.0.0.1 Feb 4 '14 at 20:28
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We have $C_{A}(x) = (x-1)^{5}(x-2)^{2}$ and $m_{A}(x) = (x-1)^{3}(x-2)^{2}$. Write $B = A^{2} + A + I$. Then $Av = \lambda v \implies Bv = (\lambda^{2} + \lambda + 1)v$. This equation tells us what the eigenvalues of $B$ look like, but also more: $ \ker(A-\lambda I) \subset \ker \left (B - (\lambda^{2} + \lambda + 1)I\right)$ and the reverse inclusion also holds. Therefore, the structure of $J_{A}$ is preserved. Thus, $C_{B}(x) = (x-3)^{5}(x-7)^{2}$ and $m_{B}(x) = (x-3)^{3}(x-7)^{2}$.

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