Complex Numbers form Vector Space over Themselves

Theorem

The set of complex numbers $\C$, with the operations of addition and multiplication, forms a vector space.

Proof

Let the field of complex numbers be denoted $\struct {\C, +, \times}$.

By Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is an abelian group.

From Complex Multiplication Distributes over Addition:

\(\ds \forall x, y, z \in \C: \, \)

\(\ds x \times \paren {y + z}\)

\(=\)

\(\ds x \times y + x \times z\)

\(\ds \paren {y + z} \times x\)

\(=\)

\(\ds y \times x + z \times x\)

From Complex Multiplication is Associative:

$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$

From Complex Multiplication Identity is One:

$\forall x \in \C: 1 \times x = x$

Therefore $\struct {\C, +, \times}$ forms a vector space.

$\blacksquare$

Also see

• Properties of Complex Numbers

• Complex Numbers form Vector Space over Reals

Sources

• 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 $2$