# $P_B(x)=P_A(x^k)$, what is the relationship between root of $P_B(x)$ and $P_A(x)$?

I have $A\in \mathbb{R}^{n\times n}$, and $B\in$: $$B=\begin{pmatrix} 0 & I_n&0&\cdots&0\\0&0&I_n&\cdots&0\\ \vdots&\vdots& \vdots&\ddots &\vdots\\0&0&0&\cdots&I_n\\ A & 0 &0 &\cdots &0 \end{pmatrix}_{(kn)\times(kn)}\ B\in\mathbb{R}^{kn\times kn}$$ So, $P_B(x)=P_A(x^k)$, $x\in \mathbb{C}$ (see proof here Permutation and characteristic polynomial of a matrix first answer)

Q: what is the relationship between $P_B(x)$ and $P_A(x)$ if term of their roots (arguments) or in term of eigenvalues of $A$ and $B$ ?

Let $y$ be a root of $P_A$. Let $x*$ be a $k$th root of $y$. Then

\begin{align} 0 &= P_A(y) &\mid \text{y$root of$P_A$} \\ &= P_A((x*)^k) &\mid \text{$x*k$th root of$y} \\ &= P_B(x*) &\mid \text{ given} \end{align}

. Thus, the $k$th roots of an arbitrary root of $P_A$ are roots of $P_B$.

Similarly, let $x*$ be an arbitrary root of $P_B$. Then

\begin{align} 0 &= P_B(x*) \\ &= P_A((x*)^k) \end{align}

. Thus, the $k$th potence of an arbitrary root of $P_B$ is a root of $P_A$.

• It's very helpful, any idea about relationship in term of arguments of roots? – zazar Sep 17 '18 at 16:59
• Hint: If $z \neq 0$ and $n$ is any integer, then $\operatorname{Arg}(z^n) \equiv n\operatorname{Arg}(z) \mod [-\pi, \pi]$ (source) – user7427029 Sep 17 '18 at 21:04