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Suppose $p(x) \in R[x]$ is an irreducible polynomial of degree $n$ over a commutative ring $R$. Let $A$ be the companion matrix of $p(x)$, and $I$ be the $n$ by $n$ identity matrix.

Now define

$M = \begin{bmatrix} A & I & & 0\\ & A & \ddots & \\ & & \ddots & I \\ 0 & & & A \end{bmatrix}$

where there are $k$ copies of $A$.

Is it true that the minimal polynomial of $M$ is equal to $[p(x)]^k$? (If so, how to prove?)

Thanks!

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  • $\begingroup$ Hint: try induction and en.wikipedia.org/wiki/Determinant#Block_matrices $\endgroup$ – J.G Oct 12 '19 at 0:11
  • $\begingroup$ Hi J.G, thanks for your reply. Suppose we removed the $I$ matrices from the block matrix $M$ and replaced them with $0$'s. Then $M - xI$ would have determinant $[p(x)]^k$, but its minimal polynomial would be $p(x)$. So, I think maybe we need more than what you're suggesting? (If I've inferred correctly what you're getting at.) $\endgroup$ – Jeremiah Goertz Oct 12 '19 at 0:31
  • $\begingroup$ Oh whoops, I'm so sorry, I was reading quickly and completely misread the question... $\endgroup$ – J.G Oct 12 '19 at 0:34
  • $\begingroup$ Just edited this to add the condition that p(x) is irreducible, which I believe is needed. $\endgroup$ – Jeremiah Goertz Oct 12 '19 at 1:50
  • $\begingroup$ I just found an answer here: math.stackexchange.com/questions/1570178/… This is can be generalized to solve the problem. Thanks to everyone who took a look! $\endgroup$ – Jeremiah Goertz Oct 12 '19 at 4:17
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The solution given here

Minimal polynomial of a matrix of matrices.

generalizes as a solution to my question. (If $q$ is a polynomial, note that entries of $q(M)$ are of the form $\frac{q^{(j)}(A)}{j!}$.)

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