3
$\begingroup$

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 $T$?

$\endgroup$
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.