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.
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 $m_T(x)=f(x)$.