# Eigenpolynomial of a linear operator

Let $V$ be a $n$-dimensional vector space over a field $F$, let $A\in \text{End}(V)$.

Let $q\in F[x]$ be an irreducible polynomial and $$V_q:=\{x∈V:q(A)x=0\}.$$

I wish to prove that $V_q$ is not trivial iff $q$ divides the characteristic polynomial $\chi$ of the operator $A$.

Let $\mu(x)$ be the minimal polynomial of $A$ (i.e. the polynomial of least degree such that $\mu(A)=0$). Suppose $q$ is irreducible. Then $q$ divides $\chi$ if and only if $q$ divides $\mu$ (since $\mu$ and $\chi$ have the same irreducible factors).

Now suppose $\mu(x)=p(x)q(x)$ and let $V_q=\ker q(A)$. Suppose $V_q$ is trivial, then since $V$ is finite dimensional, $q(A)$ is an isomorphism. Also since $\mu(A)=0$ and $q(A)$ is an isomorphism, one finds that $p(A)=0$. Due to minimality of $\mu$ this cannot happen unless $\deg q(x)=0$.

Since $q$ is irreducible, the fact that $q$ doesn't divide $\chi$ is equivalent to the fact that $\text{gcd}(q,\chi) = 1$. Thus Bezout's Theorem shows that we can find two polynomials $p_1$ and $p_2$ such that $$p_1 q + p_2 \chi = 1.$$

This implies that $$\forall x\in V,\; [p_1(A) q(A) + p_2(A) \chi(A)](x) = \text{Id}_A(x) = x.$$

But $\chi(A) = 0$. Hence $$\forall x\in V,\; p_1(A) q(A)(x) = x.$$

Thus if $x\ne 0$, $q(A)(x)$ can't be $0$ - otherwise, we would obtain $x = p_1(A)q(A)(x) = p_1(A)(0) = 0.$

Finally, we have shown that $V_q$ is trivial if $q$ doesn't divide $\chi$.

(Of course, it's only half of the solution. For the other part, I think Frobenius decomposition can help but I was unable to conclude.)

For every vector $v \in V$ there exists a unique monic polynomial $p$ of minimal degree such that $p(A)v = 0$.

If $f(A)v= 0$ then $p \mid f$. There exist unique $g, h$ such that $f = gp + h$ with $\deg h < \deg p$. In particular

$$0 = f(A)v = g(A)p(A)v + h(A)v = h(A)v \implies h = 0 \implies p \mid f$$ by minimality of $p$.

We can assume that $q$ is monic.

Assume $V_q \ne \{0\}$ and let $v \in V_q, v \ne 0$.

Let $p$ be the unique monic polynomial of minimal degree such that $p(A)v = 0$. Hence $p \mid q$ and since $q$ is irreducible, we conclude $p = q$.

Since $\chi(A) = 0$ by Hamilton-Cayley, we have $q \mid \chi$.

Conversely, assume that $V_q = \{0\}$. Let $\overline{F}$ be an algebraic closure of $F$. Then $q(A)$ is invertible so $0 \notin \sigma_{\overline{F}}(q(A)) = q(\sigma_{\overline{F}}(A))$ where $\sigma_{\overline{F}}$ is the spectrum of these maps in $\overline{F}$.

Let $\lambda \in \overline{F}$ be a zero of $q$. Then $\lambda \notin \sigma_{\overline{F}}(A)$ so $\chi(\lambda) \ne 0$. Hence $q \not\mid \chi$.