Complex Numbers form Vector Space over Reals

Theorem

Let $\R$ be the set of real numbers.

Let $\C$ be the set of complex numbers.

Then the $\R$-module $\C$ is a vector space.

Proof

Recall that Real Numbers form Field.

Thus by definition, $\R$ is also a division ring.

Thus we only need to show that $\R$-module $\C$ is a unitary module, by demonstrating the module properties:

$\forall x, y, \in \C, \forall \lambda, \mu \in \R$:

$(1): \quad \lambda \paren {x + y} = \paren {\lambda x} + \paren {\lambda y}$

$(2): \quad \paren {\lambda + \mu} x = \paren {\lambda x} + \paren {\mu x}$

$(3): \quad \paren {\lambda \mu} x = \lambda \paren {\mu x}$

$(4): \quad 1 x = x$

As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \C$.

Thus from Complex Multiplication Distributes over Addition, $(1)$ and $(2)$ immediately follow.

$(3)$ follows from Complex Multiplication is Associative.

$(4)$ follows from Complex Multiplication Identity is One, as $1 + 0 i$ is the unity of $\C$.

$\blacksquare$

Also see

• Properties of Complex Numbers

• Complex Numbers form Vector Space over Themselves

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 26$: Example $26.2$
• 1974: Robert Gilmore: Lie Groups, Lie Algebras and Some of their Applications ... (previous) ... (next): Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $2$