Page 82 - Linear Algebra Done Right

Chapter 4. Polynomials
68
4.8
has a unique factorization (except for the order of the factors) of the
form Corollary: If p ∈P(C) is a nonconstant polynomial, then p
4.9 p(z) = c(z − λ 1 )...(z − λ m ),
where c, λ 1 ,...,λ m ∈ C.
Proof: Let p ∈P(C) and let m denote the degree of p. We will use
induction on m.If m = 1, then clearly the desired factorization exists
and is unique. So assume that m> 1 and that the desired factorization
exists and is unique for all polynomials of degree m − 1.
First we will show that the desired factorization of p exists. By the
fundamental theorem of algebra (4.7), p has a root λ. By 4.1, there is a
polynomial q with degree m − 1 such that

p(z) = (z − λ)q(z)

for all z ∈ C. Our induction hypothesis implies that q has the desired
factorization, which when plugged into the equation above gives the
desired factorization of p.
Now we turn to the question of uniqueness. Clearly c is uniquely
determined by 4.9—it must equal the coefficient of z m in p. So we need
only show that except for the order, there is only one way to choose
λ 1 ,...,λ m .If

(z − λ 1 )...(z − λ m ) = (z − τ 1 )...(z − τ m )

for all z ∈ C, then because the left side of the equation above equals 0
when z = λ 1 , one of the τ’s on the right side must equal λ 1 . Relabeling,
we can assume that τ 1 = λ 1 . Now for z = λ 1 , we can divide both sides
of the equation above by z − λ 1 , getting

(z − λ 2 )...(z − λ m ) = (z − τ 2 )...(z − τ m )

for all z ∈ C except possibly z = λ 1 . Actually the equation above
must hold for all z ∈ C because otherwise by subtracting the right side
from the left side we would get a nonzero polynomial that has infinitely
many roots. The equation above and our induction hypothesis imply
that except for the order, the λ’s are the same as the τ’s, completing
the proof of the uniqueness.