# Linear Algebra: #16 Complex Numbers

On the other hand, looking at the characteristic polynomial, namely x

^{2}− 2x cos θ +1 in the previous example, we see that in the case θ = ±π this reduces to x

^{2}+ 1. And in the realm of the complex numbers ℂ, this equation

*does*have zeros, namely ±i. Therefore we have the seemingly bizarre situation that a “complex” rotation through a quarter of a circle has vectors which are mapped back onto themselves (multiplied by plus or minus the “imaginary” number i). But there is no need for panic here! We need not follow the example of numerous famous physicists of the past, declaring the physical world to be “paradoxical”, “beyond human understanding”, etc. No. What we have here is a purely

*algebraic*result using the abstract mathematical construction of the complex numbers which, in this form, has nothing to do with rotations of

*real physical*space!

So let us forget physical intuition and simply enjoy thinking about the artiﬁcial mathematical game of extending the system of real numbers to the complex numbers. I assume that you all know that the set of complex numbers ℂ can be thought of as being the set of numbers of the form x + yi, where x and y are elements of the real numbers ℜ and i is an abstract symbol, introduced as a “solution” to the equation x

^{2}+ 1 = 0. Thus i

^{2}= −1. Furthermore, the set of numbers of the form x + 0 · i can be identiﬁed simply with x, and so we have an embedding ℜ ⊂ ℂ. The rules of addition and multiplication in ℂ are

Let z = x +yi be some complex number. Then the

*absolute value*of z is deﬁned to be the (non-negative) real number |z| = √(x

^{2}+ y

^{2}). The

*complex conjugate*of z is z = x − yi. Therefore |z| = √(zz).

It is a simple exercise to show that ℂ is a ﬁeld. The main result — called (in German) the Hauptsatz der Algebra — is that ℂ is an

*algebraically closed*ﬁeld. That is, let ℂ[z] be the set of all polynomials with complex numbers as coefficients. Thus, for P(z) ∈ ℂ[z] we can write P(z) = c

_{n}z

^{n}+ c

_{n}

_{−1}z

^{n}

^{−1}+ · · · + c

_{1}z + c

_{0}, where c

_{j}∈ ℂ, for all j = 0, . . . , n. Then we have:

**Theorem 44 (Hauptsatz der Algebra)**Let P(z) ∈ ℂ[z] be an arbitrary polynomial with complex coefficients. Then P has a zero in ℂ. That is, there exists some λ ∈ ℂ with P(λ) = 0.

The theory of complex numbers (Funktionentheorie in German) is an extremely interesting and pleasant subject. Complex analysis is quite different from the real analysis.

**Theorem 45**Every complex polynomial can be completely factored into linear factors. That is, for each P(z) ∈ ℂ[z] of degree n, there exist n complex numbers (perhaps not all different) λ

_{1}, . . . , λ

_{n}, and a further complex number c, such that

P(z) = c(λ

_{1}− z) · · · (λ_{n}− z).*Proof*

Given P(z), theorem 44 tells us that there exists some λ

_{1}∈ ℂ, such that P(λ

_{1}) = 0. Let us therefore divide the polynomial P(z) by the polynomial (λ

_{1}− z). We obtain

P(z) = (λ

_{1}− z) · Q(z) + R(z),where both Q(z) and R(z) are polynomials in ℂ[z]. However, the degree of R(z) is less than the degree of the divisor, namely (λ

_{1}− z), which is 1. That is, R(z) must be a polynomial of degree zero, i.e. R(z) = r ∈ ℂ, a constant. But what is r? If we put λ

_{1}into our equation, we obtain

0 = P((λ

_{1}) = (λ_{1}− λ_{1})Q(z) + r = 0 + r.Therefore r = 0, and so

P(z) = (λ

_{1}− z)Q(z),where Q(z) must be a polynomial of degree n−1. Therefore we apply our argument in turn to Q(z), again reducing the degree, and in the end, we obtain our factorization into linear factors.

So the consequence is: let

**V**be a vector space over the ﬁeld of complex numbers ℂ. Then

*every*linear mapping f :

**V**→

**V**has at least one eigenvalue, and thus at least one eigenvector.

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