$p(x)$ divides the minimal polynomial iff $\exists v\ne 0: p(T)(v)=0$

Question

Let $V$, a finite dimensional space. Let $T:V\to V$ a linear transformation. Show that $p(x)$, an irreducible polynomial divides $m_T$ (The minimal polynomial of $T$) iff there is a $V\ni v \ne 0$ such that $P(T)(v)=0$.

We've learn that if $p(x)$ is an irreducible polynomial it must have the form $x-\alpha$ or $x^2+bx+c$ where $b^2-4c < 0$.

So lets split into cases:

1. $p(x)=x-\lambda$: $$p(x)|m_T \iff m_T = (x-\lambda)q(x) \iff \exists v\ne 0: p(T)(v) = 0 \iff \exists v\ne 0(T-\lambda)(v)=0$$

What should I do in the case where $p(x)=x^2+bx+c$?

Answer 1

Hint: Let $K$ your base field. If your polynomial $p$ does not divide the minimal polynomial $M$ of $T$, then as it is irreductible it is prime to $M$. Then use Bézout's theorem: There exists $U,V\in K[x]$ such that $U(x)M(x)+V(x)p(x)=1$.