2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\frac{q^{(j)}(A)}{j!}$.)

$\endgroup$

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.