Linear Algebra, Theory And Applications, 2012a

18 PRELIMINARIES
Example 1.5.6 Find the three cube roots of i.
First note that i =1 ( cos ( ) (
π
2 + i sin π
))
2 . Using the formula in the proof of the above
corollary, the cube roots of i are
( ( ) ( ))
(π/2) + 2lπ (π/2) + 2lπ
1 cos
+ i sin
3
3
where l =0, 1, 2. Therefore, the roots are
( π
) ( π
) ( ) ( )
5 5
cos + i sin , cos
6 6 6 π + i sin
6 π ,
and
( ) ( )
3 3
cos
2 π + i sin
2 π .
Thus the cube roots of i are √ 3
2 + i ( )
1
2 ,
− √ 3
2
+ i ( 1
2)
, and −i.
The ability to find k th roots can also be used to factor some polynomials.
Example 1.5.7 Factor the polynomial x 3 − 27.
First find the cube roots ( of 27. By the above procedure using De Moivre’s theorem,
−1
these cube roots are 3, 3
2 + i √ ) (
3
−1
2
, and 3
2 − i √ )
3
2
. Therefore, x 3 +27=
Note also
( ( √ )) ( ( √ ))
−1 3
(x − 3) x − 3
2 + i −1 3
x − 3
2
2 − i .
2
( (
−1
x − 3
2 + i √ )) ( (
3
−1
2
x − 3
2 − i √ ))
3
2
= x 2 +3x + 9 and so
x 3 − 27 = (x − 3) ( x 2 +3x +9 )
where the quadratic polynomial, x 2 +3x + 9 cannot be factored without using complex
numbers.
The real and complex numbers both are fields satisfying the axioms on Page 13 and it is
usually one of these two fields which is used in linear algebra. The numbers are often called
scalars. However, it turns out that all algebraic notions work for any field and there are
many others. For this reason, I will often refer to the field of scalars as F although F will
usually be either the real or complex numbers. If there is any doubt, assume it is the field
of complex numbers which is meant. The reason the complex numbers are so significant in
linear algebra is that they are algebraically complete. This means that every polynomial
∑ n
k=0 a kz k ,n≥ 1,a n ≠0, having coefficients a k in C has a root in in C.
Later in the book, proofs of the fundamental theorem of algebra are given. However, here
is a simple explanation of why you should believe this theorem. The issue is whether there
exists z ∈ C such that p (z) =0forp (z) a polynomial having coefficients in C. Dividing by
the leading coefficient, we can assume that p (z) is of the form
p (z) =z n + a n−1 z n−1 + ···+ a 1 z + a 0 , a 0 ≠0.
If a 0 =0, there is nothing to prove. Denote by C r thecircleofradiusr in the complex plane
which is centered at 0. Then if r is sufficiently large and |z| = r, thetermz n is far larger
than the rest of the polynomial. Thus, for r large enough, A r = {p (z) :z ∈ C r } describes
a closed curve which misses the inside of some circle having 0 as its center. Now shrink r.