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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    $\begingroup$ 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 $A$). $\endgroup$ – Daniil Jan 4 '12 at 18:45
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    $\begingroup$ The minimal polynomial divides any polynomial annihilating $T$. $\endgroup$ – Pierre-Yves Gaillard Jan 4 '12 at 18:50
  • $\begingroup$ @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 $m_T|\chi_T$ $\endgroup$ – Freeman Jan 4 '12 at 18:58
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    $\begingroup$ 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.] $\endgroup$ – Pierre-Yves Gaillard Jan 4 '12 at 19:17
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    $\begingroup$ 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. $\endgroup$ – Pierre-Yves Gaillard Jan 4 '12 at 19:30

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.


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