# Minimal polynomial of composite matrix

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!

• Hint: try induction and en.wikipedia.org/wiki/Determinant#Block_matrices – J.G Oct 12 '19 at 0:11
• 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.) – Jeremiah Goertz Oct 12 '19 at 0:31
• Oh whoops, I'm so sorry, I was reading quickly and completely misread the question... – J.G Oct 12 '19 at 0:34
• Just edited this to add the condition that p(x) is irreducible, which I believe is needed. – Jeremiah Goertz Oct 12 '19 at 1:50
• 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! – Jeremiah Goertz Oct 12 '19 at 4:17

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