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Let $V$ be a $K$ vector space and $f: V \to V$ $K$-linear. Let $v \in V \setminus \{0\}$ and $p \in K[X]$ with $p(f)(v) = 0$. How does one proof that $p$ and the characteristic polynomial $\chi_f$ are not coprime?


Ideas: As the roots of $\chi_f$ are the eigenvalues of $f$, there must be an eigenvalue $\lambda$ such chat $X - \lambda$ divides $p$. By Cayley-Hamilton, we have $\chi_f(f)(v) = 0$. But I have no idea how to get any knowledge about $p$ via $p(f)(v) = 0$.

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Define the set $I = \{ g \in K[x] \mid g(f)(v) = 0 \}$, i.e. the set of polynomials in $f$ that annihilate $v$. Check that it's an ideal.

Recall that $K[x]$ is a principal ideal domain when $K$ is a field, so $I$ is principally generated. We'll call its generator the minimal polynomial (of $f$ with respect to $v$).

Of course $p, \chi_f$ are both elements of $I$, so the minimal polynomial (of $f$ with respect to $v$) divides them both.

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