Page 80 - Linear Algebra Done Right

Chapter 4. Polynomials
66
4.4
Corollary: Suppose a 0 ,...,a m ∈ F.If
2
a 0 + a 1 z + a 2 z +· · ·+ a m z
for all z ∈ F, then a 0 =· · ·= a m = 0. m = 0
2
Proof: Suppose a 0 +a 1 z+a 2 z +· · ·+a m z m equals 0 for all z ∈ F.
By 4.3, no nonnegative integer can be the degree of this polynomial.
Thus all the coefficients equal 0.
m
The corollary above implies that (1,z,...,z ) is linearly indepen-
dent in P(F) for every nonnegative integer m. We had noted this earlier
(in Chapter 2), but now we have a complete proof. This linear indepen-
dence implies that each polynomial can be represented in only one way
j
as a linear combination of functions of the form z . In particular, the
degree of a polynomial is unique.
If p and q are nonnegative integers, with p = 0, then there exist
nonnegative integers s and r such that

q = sp + r.

and r< p. Think of dividing q by p, getting s with remainder r. Our
next task is to prove an analogous result for polynomials.
Let deg p denote the degree of a polynomial p. The next result is
often called the division algorithm, though as stated here it is not really
an algorithm, just a useful lemma.

Think of 4.6 as giving 4.5 Division Algorithm: Suppose p, q ∈P(F), with p = 0. Then
the remainder r when there exist polynomials s, r ∈P(F) such that
q is divided by p.
4.6 q = sp + r
and deg r< deg p.

Proof: Choose s ∈P(F) such that q − sp has degree as small as
possible. Let r = q − sp. Thus 4.6 holds, and all that remains is to
show that deg r< deg p. Suppose that deg r ≥ deg p.If c ∈ F and j is
a nonnegative integer, then

j
j
q − (s + cz )p = r − cz p.
Choose j and c so that the polynomial on the right side of this equation
has degree less than deg r (specifically, take j = deg r −deg p and then