# linear algebra – q (T) = 0 implies that q is a multiple of the minimum polynomial

I read Sheldon Axler Linear Algebra Done Right Third Edition. There is evidence on page 264. I don't understand any of the equality.

Sentence: Accept $$T in mathcal {L} (V)$$ and $$q in mathcal {P} ( mathbb {F})$$. Then $$q (T) = 0$$ then and only if $$q$$ is a polynomial multiple of the minimal polynomial of $$T$$.

Proof:
To let $$p$$ denote the minimal polynomial of $$T$$.
Accept $$q (T) = 0$$.
According to the division algorithm for polynomials, there are polynomials $$s, r in mathcal {P} ( mathbb {F})$$ so that $$q = ps + r$$ and degrees $$r$$ <Degrees $$s$$.

We have $$0 = q (T) = p (T) s (T) + r (T) = r (T)$$.

The above equation implies that r = 0 (otherwise divided $$r$$ its highest degree coefficient would produce a monic polynomial which, when applied to $$T$$ returns 0; This polynomial would be less than the minimal polynomial, which would be a contradiction.

My question is:
How do we know? $$p (T) s (T) + r (T) = r (T)$$?