Complex Numbers form Vector Space

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

First note that $\R$, being a field, is also a division ring.

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

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

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

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

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

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

As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \C$ and so $(1)$ and $(2)$ immediately follow from the fact that the complex numbers form a field.

So multiplication distributes over addition in $\C$.

$(3)$ follows from the fact that multiplication is associative on $\C$, again because $\C$ is a field.

$(4)$ follows as $1 + 0 i$ is the unity of $\C$.

$\blacksquare$

Also see

• Properties of the Complex Numbers

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 26$: Example $26.2$