04 - Minimal Polynomial divides Annihilating Polynomials

Minimal Polynomial divides Annihilating Polynomials

If $f(A)=0$ then

$$m_A\mid f$$

where $m_A$ is the minimal polynomial

Furthermore, $m_A$ is unique

Proof - Main

By Division Algorithm for Polynomials then
There exist polynomials, $q,r$ with $\mathrm{deg}r<\mathrm{deg}{m}_A$ such that

$$f=q{m}_A+r$$

Evaluating at $A$ gives $f(A)=0=r(A)$ so as $m_A$ is minimal then

$$r=0$$

Hence $m_A$ divides $f$

Proof - Uniqueness

Suppose $m$ is another monic polynomial of minimal degree and $m(A)=0$
So by the theorem above then $m_A\mid m$

As $m$ and $m_A$ have the same degree then

$$m=a{m}_A\text{ for some }a\in \mathbf{F}$$

As both polynomials are monic then $a=1$ so $m=m_A$