Complex Numbers form Vector Space over Reals
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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
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: 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 $2$