Show that $\ker f(T) = \ker d(T)$

Question

I'm having trouble solving the following question:

Let $T$ be a linear operator on $V$ with minimal polynomial $p(x)$ over $F$. Let $f(x)$ be any polynomial over $F$ and $d(x)$ be the greatest common divisor of $f(x)$ and $p(x)$. Prove that $\ker f(T) = \ker d(T)$.

Answer 1

Since $d=\gcd(f,p)$, there are polynomials $a,b$ such that $af+bp=d$.

Therefore, $d(T)=a(T)f(T)+b(T)p(T)=a(T)f(T)$.

This implies that if $f(T)x=0$, then $d(T)x=0$. In other words, $\ker(f(T))\subset\ker(d(T))$.

The converse is trivial. Since there is $c$ such that $f=cd$, then $f(T)x=c(T)d(T)x$. Therefore, if $d(T)x=0$, this implies that $f(T)x=0$. In other words $\ker(d(T))\subset \ker(f(T))$.