Complex Numbers and Complex Vector Spaces

18 Complex Numbers and Complex Vector Spaces

So far in this course, we have only looked at vectors spaces "over $R$ ," also called "real vector spaces." In the definition of a vector space over $R$ , the scalars are real numbers. The standard $n$ -dimensional vectors space is $R^{n}$ . Linear maps are represented by matrices of real numbers.

However, the real numbers are just one example of a "field." A field is a certain kind of algebraic structure. There are other fields besides $R$ , and any field can be used in the definition of vector space in place of $R$ . In particular, we will need to use $C$ , the field of complex numbers, and vectors spaces over $C$ . The definition of field pretty much just says that addition and multiplication in the field have all the familiar properties of addition and multiplication of real numbers.

Definition: A field, $(F, +, \cdot)$ , consists of a set $F$ and two binary operators, $+$ and $\cdot$ , on $F$ satisfying

Closure under addition: for all $a, b \in F$ , $a + b \in F$ .
Addition is commutative: for all $a, b \in F$ , $a + b = b + a$ .
Addition is associative: for all $a, b \in F$ , $a + (b + c) = (a + b) + c$ .
Additive identity: there exists $0 \in F$ such that $0 \cdot a = 0$ for all $a \in F$ .
Additive inverses: for every $a \in F$ , there exists $- a \in F$ such that $a + (- a) = 0$ .
Closure under multiplication: for all $a, b \in F$ , $a \cdot b \in F$ .
Multiplication is commutative: for all $a, b \in F$ , $a \cdot b = b \cdot a$ .
Multiplication is associative: for all $a, b \in F$ , $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ .
Multiplicative identity: there exists $1 \in F$ such that $1 \cdot a = a$ for all $a \in F$ .
Multiplicative inverses: for any $a \in F$ , if $a \neq 0$ , there exists $a^{- 1} \in F$ such that $a \cdot a^{- 1} = 1$ .
Multiplication distributes over addition: for all $a, b, c \in F$ , $c \cdot (a + b) = (c \cdot a) + (c \cdot b)$ .
$1 \neq 0$ .

The set $Q$ of rational numbers is a field that is a subset of $R$ . $Q$ is a subfield of $R$ becasue it uses the addition and multiplication operations that it inherits from $R$ . $R$ actually has many other subfields, such as ${a + b \sqrt{2} | a, b \in Q}$ . There are even finite fields. For example the set $Z_{p} = {0, 1, 2, \dots, p - 1}$ , where $p$ is any prime number, becomes a field if addition and multiplication are defined "mod p". (This just means that in $Z_{p}$ , $a$ plus $b$ means the remainder when the usual $a + b$ is divided by $p$ , and similarly for $a$ times $b$ .) However, the only field that we will actually use is the field of complex numbers.

A complex number can be written in the form $a + b i$ where $a$ and $b$ are real numbers and $i$ is used to represent the constant $\sqrt{- 1}$ . A real number $a$ is also a complex number since it can be written in the form $a + 0 \cdot i$ . So $R$ is a subset of $C$ . But to make $C$ into a field, we have to define addition and multiplication of complex numbers. This is done as follows: $\begin{aligned} (a + b i) + (c + d i) & = (a + b) + (c + d) i \\ (a + b i) \cdot (c + d i) & = a c + a d i + b c i + b d i^{2} \\ = a c + a d i + b c i - b d \\ = (a c - b d) + (a d + b c) i \end{aligned}$ It can be shown that with these definitions, $(C, +, \cdot)$ is a field. The multiplicative inverse of a non-zero complex number $a + b i$ is given by $\frac{1}{a + b i} = \frac{a}{a^{2} + b^{2}} - \frac{b}{a^{2} + b^{2}} i$ . More generally, we can compute a quotient as $\frac{a + b i}{c + d i} = \frac{a + b i}{c + d i} \cdot \frac{c - d i}{c - d i} = \frac{(a + b i) (c - d i)}{c^{2} + d^{2}} = \frac{a c + b d}{c^{2} + d^{2}} + \frac{a d - b c}{c^{2} + d^{2}} i$

The complex number $a + b i$ can be visualized as the point $(a, b)$ in the plane. The complex plane consists of all the complex numbers, visualized in this way. The horizontal axis of the plane represents the real numbers and is called the real axis. The vertical axis consists of all multiples of $i$ and is called the imaginary axis. Note that this identifies $C$ with $R^{2}$ , and in fact $C$ can be considered to be a real vector space of dimension two, with scalar multiplication $r \cdot (a + b i) = r a + r b i$ for $r \in R$ .

If $z = a + b i$ is a complex number, then the norm (or length or absolute value) of $z$ is defined to be $| z | = | a + b i | = \sqrt{a^{2} + b^{2}}$ . Note that the norm is a non-negative real number. The conjugate of $z$ is defined to be $\overset{―}{z} = \overset{―}{a + b i} = a - b i$ . Note that $z \in R$ if and only if $z = \overset{―}{z}$ . Also, $| z |^{2} = z \cdot \overset{―}{z}$ .

Recall that it is possible for a polynomial $p (x)$ of degree two or higher, with real coefficients, to have no roots. That is, the equation $p (x) = 0$ might have no solutions. The simplest example is $p (x) = x^{2} + 1$ . There is no real number $x$ such that $x^{2} + 1 = 0$ . However, the complex numbers $i$ and $- i$ are complex roots of this equation. Note that $x^{2} + 1$ cannot be factored if we only have real numbers to work with. We say that $x^{2} + 1$ is "irreducible over the real numbers." But if we are using complex numbers, we can factor $x^{2} + 1 = (x - i) (x + i)$ . One of the most important and surprising facts about complex numbers — and the main reason why we need them in this course — is the following theorem:

Fundamental Theorem of Algebra: Any non-constant polynomial with complex coefficients has a root in the complex numbers. Any complex polynomial of degree $n > 0$ factors into a product of $n$ linear factors.

To define vector spaces over a field $F$ , we can use exactly the same definition as for vector spaces over $R$ , except that scalars are elements of $F$ instead of elements of $R$ . We will only be interested in vector spaces over $C$ , which are also called complex vector spaces. Almost everything that we did with real vector spaces could have been done with complex vector spaces. The only exception is things involving inner product (including length, orthogonality, and angles), which has to be defined differently for complex column vectors).

Thus, we define column vectors, row vectors, matrices, row operations, echelon form, linear combinations, matrices, matrix multiplication, homomorphisms, and determinants using complex numbers instead of real numbers.

The standard $n$ -dimensional vector space is $C^{n}$ , the vector space of column vectors where the entries can be complex numbers. The standard basis for $C^{n}$ is the same as the standard basis for $R^{n}$ , $E_{n} = ⟨ {\vec{e}}_{1}, {\vec{e}}_{2}, \dots, {\vec{e}}_{n} ⟩$ . Any $n$ -dimensional complex vector space is isomorphic to $C^{n}$ . We can redefine $P_{n}$ to be the complex vector space of polynomials with complex coefficients and degree less than or equal to $n$ , and we then have that $P_{n}$ is isomorphic to $C^{n + 1}$ . Similarly, $M_{m \times n}$ now represents $m \times n$ matrices with complex entries, a complex vector space of dimension $m n$ . Any $m \times n$ matrix $A$ of complex numbers defines a homomorphism from $C^{n}$ to $C^{m}$ by sending $\vec{v}$ to $A \vec{v}$ .

Note that the determinant of an $n \times n$ complex matrix will be a complex number. It is still true that the determinant is zero if and only if the matrix is singular.

In short, nothing much changes except that multiplication and division become harder. The reason for making the change from real to complex vector spaces will become clear as we study eigenvalues and eigenvectors.

(back to contents)