Page 86 - Linear Algebra Done Right

Chapter 4. Polynomials
72
and (x − λ) is a term in the factorization of p as an element of P(C).
¯
We are guaranteed by 4.10 that (x − λ) also appears as a term in the
factorization, but 4.10 does not state that these two factors appear
the same number of times, as needed to make the idea above work.
However, all is well. We can write
¯
p(x) = (x − λ)(x − λ)q(x)
2 2
= x − 2(Re λ)x +|λ| q(x)
for some polynomial q ∈P(C) with degree two less than the degree
of p. If we can prove that q has real coefficients, then, by using induc-
tion on the degree of p, we can conclude that (x − λ) appears in the
¯
factorization of p exactly as many times as (x − λ).
To prove that q has real coefficients, we solve the equation above
Here we are not for q, getting
dividing by 0 because p(x)
q(x) = 2 2
the roots of x − 2(Re λ)x +|λ|
2 2
x − 2(Re λ)x +|λ| for all x ∈ R. The equation above implies that q(x) ∈ R for all x ∈ R.
¯
are λ and λ, neither of Writing
which is real. q(x) = a 0 + a 1 x +· · ·+ a n−2 x n−2 ,
where a 0 ,...,a n−2 ∈ C, we thus have
0 = Im q(x) = (Im a 0 ) + (Im a 1 )x +· · ·+ (Im a n−2 )x n−2

for all x ∈ R. This implies that Im a 0 ,..., Im a n−2 all equal 0 (by 4.4).
Thus all the coefficients of q are real, as desired, and hence the desired
factorization exists.
Now we turn to the question of uniqueness of our factorization. A
2
2
factor of p of the form x +αx+β with α < 4β can be uniquely written
¯
as (x − λ)(x − λ) with λ ∈ C. A moment’s thought shows that two
different factorizations of p as an element of P(R) would lead to two
different factorizations of p as an element of P(C), contradicting 4.8.