Complex Vector Space

What a Complex Number Contains

The imaginary unit $i$ satisfies $i^2=-1$ . A complex number combines a real and an imaginary coordinate, so it can represent both scaling and rotation in one scalar.

The set of complex numbers is defined as $\mathbb{C} := \{z = x + iy : x, y \in \mathbb{R}\}$ . In every complex number $z = x + iy$ , the part $x = \text{Re}(z)$ is called the $real part$ and the part $y = \text{Im}(z)$ is called the $imaginary part$ .

Addition, subtraction, multiplication, and division by a nonzero number stay inside $\mathbb C$ . With these operations, $(\mathbb{C}, +, \cdot)$ is a $field$ , so its elements can serve as scalars for a vector space.

Conjugate and Modulus

For $z=x+iy$ , the $complex conjugate$ is $\overline z=x-iy$ . Since $z\overline z=x^2+y^2$ , conjugation gives both the magnitude and, when $z\neq0$ , the reciprocal $z^{-1}=\overline z/|z|^2$ .

The $modulus$ or absolute value of a complex number is calculated with the formula:

|z| = \sqrt{z \cdot \overline{z}} = \sqrt{x^2 + y^2}

This modulus gives the "distance" of the complex number from the origin in the complex plane.

For $z=3+4i$ , conjugation changes the sign of the imaginary coordinate while preserving the distance from zero. Follow the two segments in the diagram.

\overline z=3-4i

Conjugation Reflects across the Real Axis

Horizontal coordinate: real part. Vertical coordinate: imaginary part. The points

3+4i

and

3-4i

are equally far from zero.

$x$: $[0, 3]$
$y$: $[-4, 4]$

Fundamental Theorem of Algebra

The fundamental theorem of algebra says that every nonconstant polynomial with complex coefficients has a complex root.

Let $p$ be a nonconstant polynomial with complex input $z$ . The notation $z\in\mathbb{C}$ means that $z$ can be any complex number. We write the polynomial as:

\begin{aligned} p(z) &= a_n z^n + a_{n-1} z^{n-1} + \cdots \\ &\quad{}+ a_1 z + a_0 \end{aligned}

If $n \geq 1$ and the complex coefficients $a_k \in \mathbb{C}$ for $k = 0, 1, \ldots, n$ satisfy $a_n \neq 0$ , then $p$ has at least one complex root. In other words, some $z_* \in \mathbb{C}$ satisfies $p(z_*) = 0$ .

Applying the theorem repeatedly shows that a degree- $n$ polynomial factors completely over $\mathbb C$ :

p(z)=a_n\prod_{j=1}^n(z-z_j)

It has exactly $n$ complex roots when $multiplicities are counted$ . The qualification "nonconstant" matters: a nonzero constant polynomial has no root.

Definition and Axioms of Complex Vector Space

A complex vector space uses numbers from $\mathbb C$ as scalars. The operations must be closed in the stated sets and obey the same algebraic compatibility rules as a real vector space.

Addition on a set $V$ is defined by

+ : V \times V \to V : (x, y) \mapsto x + y

Scalar multiplication on $V$ is defined by

\cdot : \mathbb{C} \times V \to V : (\lambda, x) \mapsto \lambda \cdot x

The set $V$ with these two operations is a $complex vector space$ if it satisfies the following axioms:

Vector Addition Axioms

These four axioms test the addition operation on its own, so each one names the property that addition must have for every choice of vectors in $V$ .

$Associative$ : $(x + y) + z = x + (y + z)$ for all $x, y, z \in V$
$Commutative$ : $x + y = y + x$ for all $x, y \in V$
$Identity Element$ : There exists an element $0 \in V$ such that $x + 0 = x = 0 + x$ for all $x \in V$
$Inverse Element$ : For every $x \in V$ there exists an element $-x \in V$ such that $x + (-x) = 0 = (-x) + x$

Scalar Multiplication Axioms

The two axioms here concern multiplication by a complex scalar, so each one keeps the scalar and the vector in their separate roles.

$Associative multiplication$ : $(\lambda \mu) \cdot x = \lambda \cdot (\mu \cdot x)$ for all $\lambda, \mu \in \mathbb{C}, x \in V$
$Unit element$ : $1 \cdot x = x$ for all $x \in V$

Distributive Axioms

The two distributive laws connect scalar multiplication with vector addition and scalar addition. They allow a sum to be split before or after scaling.

$Distributive over vector addition$ : $\lambda \cdot (x + y) = \lambda \cdot x + \lambda \cdot y$ for all $\lambda \in \mathbb{C}, x, y \in V$
$Distributive over scalar addition$ : $(\lambda + \mu) \cdot x = \lambda \cdot x + \mu \cdot x$ for all $\lambda, \mu \in \mathbb{C}, x \in V$

Elements $x \in V$ are called $vectors$ , and elements $\lambda \in \mathbb{C}$ are called $scalars$ .

Complex Scalar Multiplication in Two Dimensions

The set $V=\mathbb C^2$ is a complex vector space with componentwise operations. For instance,

\begin{aligned} (1+i)\begin{pmatrix}1\\-i\end{pmatrix} &= \begin{pmatrix}(1+i)\cdot1\\(1+i)(-i)\end{pmatrix} \\ &= \begin{pmatrix}1+i\\-i-i^2\end{pmatrix} = \begin{pmatrix}1+i\\1-i\end{pmatrix} \end{aligned}

$The scalar field changes how we count dimensions$ . The set $\mathbb C$ itself has basis $\{1\}$ as a complex vector space, but basis $\{1,i\}$ as a real vector space:

Same set of vectors	Allowed scalars	Dimension
$\mathbb C$	$\mathbb C$	$1$
$\mathbb C$	$\mathbb R$	$2$

Differences from Real Vector Spaces

Purely algebraic results derived from the vector-space axioms work over both $\mathbb R$ and $\mathbb C$ . Results that use order or a real-only inner product need their assumptions adjusted. For example, complex inner products use conjugation, and complex numbers have no field order compatible with their arithmetic.

The $scalar field$ also decides which maps are linear. Complex conjugation is linear over $\mathbb R$ . Over $\mathbb C$ , multiplication by $i$ shows why linearity fails: $\overline{iz}=-i\overline z$ . Always state the scalar field before discussing a basis, dimension, or linear map.