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The Cayley Hamilton theorem states for a transformation $T:V \rightarrow V$ then the characteristic equation of $T$, $X_T(x)$ has the property that $X_T(A)=0$ where A is the matrix representation of the transformation. Equivalently $X_A(A)=0$?

Can anyone explain to me why this is equivalent to $m_T|X_T$ where $m_T$ is the minimum polynomial of $T$?

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    Because of the definition of minimal polynomial. $m_A$ is minimal if it's a polynomial of the least degree such as $m_A(A) = 0$. That means that every root of $m_A$ is also a root of $\chi_A$ (and also the eigenvalue of $A$).2012-01-04
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    The minimal polynomial divides any polynomial annihilating $T$.2012-01-04
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    @Daniil: So as every root of $m_A$ is a root of $\chi_A$ and we have $m_A(A)=0$ and $\chi_A(A)=0$ implying that $m_T|\chi_T$2012-01-04
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    @Pierre-YvesGaillard: So why is that? Sorry if that's a stupid question. Is that because the minimum polynomial is unique?2012-01-04
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    It is more or less the definition of the minimal ideal (when done properly, I guess...)2012-01-04
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    @LHS: Yes. And the other way around: if $m_T|\chi_T$ then every root of $m_T$ is a root of $\chi_T$ which means that $\chi_T(T) = 0$ (since $m_T(T) = 0$).2012-01-04
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    Dear LHS: Let $K$ be the ground field. The polynomial annihilating $T$ form a nonzero ideal $I$ of $K[x]$. Any such ideal is generated by a unique monic polynomial. The minimal polynomial is by definition the monic generator of $I$. In particular, it divides any member of $I$. [I don't find @Daniil's statements fully convincing.]2012-01-04
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    Hmm, I see, I think I understand this better now. Introducing ideals does seem to piece it together quite nicely :) thanks!2012-01-04
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    Dear LHS: I suggest that you answer your own question. --- If you write a comment for me (or for anybody), please try to think of using the `@`sign.2012-01-04

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There's little to add to the comment by Pierre-Yves Gaillard here... Denoting by $K$ the ground field of vector space $V$, we observe that the set of all polynomials $p\in K[x]$ such that $p(T)=0$ is an ideal. Since $K[x]$ is a principal ideal domain, it follows that there exists $m_T\in K[x]$ (unique up to a unit in $K[x]$, that is a nonzero scalar) such that $$\{p\in K[x]:p(T)=0\}=\{p\in K[x]: m_T\text{ divides }p\}$$ Which was to be explained.