Exercise 7.3.11

for all $y\in V$. So the minimal polynomial $p(t)$ must divide $g(t)$. If the degree of $p(t)$ is less than but not equal to the degree of $g(t)$, then the solution space of the equation $p(D)(y)=0$ must contain $V$. This will make the dimension of the solution space of $p(D)(y)=0$ greater than the degree of $p(t)$. This is a contradiction to Theorem 2.32. Hence we must have $p(t)=g(t)$.