Show that $Tr(A)$ is in the spectrum of $A+I$

Question

If we have $p$ an odd prime number and the matrix $A$ $(p×p)$ with rational entries and we know that $\det(A^{p}+I)=0$, $\det(A+I)≠0$. How we can show $\operatorname{Tr}(A)$ is in the spectrum of $A+I$?
I know we need to use the fact that the minimal polynomial divides the cyclotomic polynomial which is irreductible over Q. But I dont know how to extract that Information from the determinant. Yes we can say that $\det(A^{p-1}-A^{p-2}+...-A+I)=0$. But I dont know how to use the minimal polynomial here. Can someone explain me?

Answer 1

Since $A+I$ is nonsingular but $A^p+I=(A+I)\Phi_p(-A)$ is singular, $\Phi_p(-A)$ must be singular. If $\Phi_p(-x)$ and the minimal polynomial $m(x)$ of $A$ are relatively prime, we would have $a(x)\Phi_p(-x)+b(x)m(x)=1$ for some polynomials $a,b\in\mathbb Q[x]$ and in turn $a(A)\Phi_p(-A)=I$, which is a contradiction. Hence $\Phi_p(-x)$ must share an irreducible factor with $m(x)$.

As $\Phi_p(-x)$ is already irreducible over $\mathbb Q$, it must divide $m(x)$. However, as $\deg\Phi_p(-x)<p<\deg\Phi_p(-x)^2$, $m(x)$ cannot possibly be equal to any power of $\Phi_p(-x)$. So, we must have $m(x)=(x-q)\Phi_p(-x)$ for some $q\in\mathbb Q$. But then $m(x)$ is also the characteristic polynomial of $A$.

Therefore $\operatorname{tr}(A)=$ sum of all zeroes of $m$ $=q+1$. Yet, as $q$ is an eigenvalue of $A$, $q+1$ is an eigenvalue of $A+I$. Hence the result follows.

Answer 2

Let $a_1,a_2,\dots,a_p$ be the eigenvalues of $A$, so that the characteristic polynomial of $A$ is $$ P_A(X)=(X-a_1)(X-a_2)\dots(X-a_p)\ . $$ The matrix $A+I$ has eigenvalues $a_1+1,a_2+1,\dots,a_p+1$, and no entry is zero, so $-1$ is not among $a_1,a_2,\dots,a_p$.

The matrix $A^p+I$ has eigenvalues $a_1^p+1,a_2^p+1,\dots,a_p^p+1$, and at least one entry is zero, so at least one entry among $a_1,a_2,\dots,a_p$ is a root of the polynomial $$ Q(X)= \frac{X^p+1}{X+1} =X^{p-1}-X^{p-2}+\dots+X^2-X+1\ . $$ Which is irreducible, it is the cyclotomic polynomial $\Phi_{2p}$ of order $2p$ and degree $\varphi(2p)=2p\left(1-\frac 12\right)\left(1-\frac 1p\right) =p-1$.

We may and do assume that $a_1$ is a root of $Q$. Then using the action of the corresponding Galois group, from $0=P_A(a_1)$, and $P_A\in\Bbb Q[X]$, for each Galois substitution $\sigma$ we have also $0=\sigma(0)=\sigma(\ P_A(a_1)\ )=P_A(\ \sigma(a_1)\ )$, so all conjugates $\sigma(a_1)$ are also roots of $P_A$. In other words, $Q$ divides $P_A$. The quotient $P_A/Q$ has degree one, and lives over $\Bbb Q$, let $b\in\Bbb Q\ne -1$ be its root, so we have: $$ P_A(X)=Q(X)(X-b)\ , $$ and the trace of $A$ is the sum of its eigenvalues, i.e. trace of $Q$ plus $b$, $$ \operatorname{Trace}(A)= 1+b\ . $$ Which is an eigenvalue of $I+A$, since $b$ is an eigenvalue of $A$.

$\square$

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Note: $A$ is diagonalizable, $A=SDS^{-1}$, with a base change matrix $S$, and $D$ diagonal, and on its diagonal there are the $(p-1)$ (different) roots of $Q$, the primitive (non-real) roots of order $2p$ of the unit, and the rational (real) entry $b$.