1
$\begingroup$

Let $V$ be a finite dimensional vector space and let $T:V\to V$ be a linear transformation. Suppose $f(T)=0$ and $f(x)=a(x)b(x)$ for some co-prime non-constant $a(x)$, $b(x)$. Then from Primary Decomposition Theorem, we know that $V=ker(a(x))\oplus ker(b(x))$. Prove that $f=m_T(x)$ if and only if $a(x)$ is the minimal polynomial for $T|_{ker(a(x))}$ and $b(x)$ is the minimal polynomial for $T|_{ker(b(x))}$.

The forward argument is solvable for me. However, in the reverse argument, I only deduced that $m_T(x)=a(x)$ or $m_T(x)=b(x)$ or $m_T(x)=f(x).$ I am not entirely sure how to eliminate the first two options. I guess one way is to show that neither $ker(a(x))$ and $ker(b(x))$ is trivial but I have no idea on how to do that either.

$\endgroup$
0
$\begingroup$

In the reverse argument, $m_T(x)$ must vanish $T_{|\ker{a(T)}}$, so $a(x)|m_T(x)$. Similarly, $b(x)|m_T(x)$, which entails ($a,b$ coprime) $f(x)=a(x)b(x)|m_T(x)$, whence $m_T(x)=f(x)$.

$\endgroup$

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.