Do these two matrices, related to $p(x)^n$ with $p(x)b$ irreducible, both have the same minimal polynomial? $$
H=\pmatrix{A&0&\cdots&\cdots&0\\ I&A&0&\cdots&0\\ 0&I&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&I&A},
\quad K=\pmatrix{A&0&\cdots&\cdots&0\\ U&A&0&\cdots&0\\ 0&U&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&U&A}.
$$
Let $T$ be an $nk\times nk$ nonsingular matrix with integer entries for which the minimal polynomial is $p(x)^n$, where $p(x)$ is irreducible of degree $k > 1$.  In the matrices $H$ and $K$, $A$ denotes the companion matrix of $p(x)$, $U$ is a  matrix every entry of which is $0$ except for the entry $1$ in the upper right corner, and $I$ is the  identity matrix.  Then $K$ is similar to $T$, and so $p(x)^n$  is the minimal polynomial of $K$.
$p(x)^n$ is the characteristic polynomial of both $H$ and $K$.
Is $p(x)^n$ also the minimal polynomial of $H$?
 A: Note that the similarity of matrices is preserved by field extension.  (See this.)
Here we assume that the irreducibility is considered over $\mathbb{Q}$, then roots of any irreducible polynomial are distinct. That's why $A$ is diagonalizable over $\overline{\mathbb{Q}}$. 
Thus, we can find an invertible $k\times k$ matrix $X$ such that $XAX^{-1}=D$, where 
$D$ is a diagonal matrix. 
Consider the invertible $nk\times nk$ matrix 
$$
Y=\pmatrix{X&0&\cdots&\cdots&0\\ 0&X&0&\cdots&0\\ 0&0&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&0&X}.
$$
Then we have 
$$YHY^{-1}=\pmatrix{D&0&\cdots&\cdots&0\\ I&D&0&\cdots&0\\ 0&I&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&I&D}.
$$
Rearranging the standard basis $\{ e_1,\cdots, e_{nk}\}$ by
$$\{e_1,e_{k+1},\cdots, e_{ (n-1)k+1} \\ e_2,e_{k+2}, \cdots, e_{ (n-1)k+2} \\ \cdots \\e_{k-1}, e_{2k-1}, \cdots , e_{nk-1} \}$$
We see that $YHY^{-1}$ is similar to the block diagonal matrix
consisted of $k\times k$ blocks of size $n\times n$, where diagonal blocks are 
$$
\pmatrix{ \lambda_i & 0 &\cdots &\cdots & 0 \\ 1 & \lambda_i & 0 & \cdots & 0 \\
0 & 1 & \lambda_i & \cdots & 0\\ \vdots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&1&\lambda_i}
$$
for each eigenvalues $\lambda_i$ of $A$. 
Hence, the minimal polynomial of $H$ is also $p(x)^n$. 
