# Real Vector Space is Vector Space

## Theorem

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

Then the real vector space $\R^n$ is a vector space.

## Proof

### Construction of Real Vector Space

From the definition, a vector space is a unitary module whose scalar ring is a field.

In order to call attention to the precise scope of the operators, let real addition and real multiplication be expressed as $+_\R$ and $\times_\R$ respectively.

Then we can express the field of real numbers as $\struct {\R, +_\R, \times_\R}$.

From Real Numbers under Addition form Group, $\struct {\R, +}$ is a group.

Again, in order to call attention to the precise scope of the operator, let real addition be expressed on $\struct {\R, +}$ as $+_G$.

That is, the group under consideration is $\struct {\R, +_G}$.

Consider the cartesian product:

$\ds \R^n = \prod_{i \mathop = 1}^n \struct {\R, +_G} = \underbrace {\struct {\R, +_G} \times \cdots \times \struct {\R, +_G} }_{n \text{ copies} }$

Let:

$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$

be arbitrary elements of $\R^n$.

Let $\lambda$ be an arbitrary element of $\R$.

Let $+$ be the binary operation defined on $\R^n$ as:

$\mathbf a + \mathbf b = \tuple {a_1 +_G b_1, a_2 +_G b_2, \ldots, a_n +_G b_n}$

Also let $\cdot$ be the binary operation defined on $\R \times \R^n$ as:

$\lambda \cdot \mathbf a = \tuple {\lambda \times_\R a_1, \lambda \times_\R a_2, \ldots, \lambda \times_\R a_n}$

In this context, $\lambda \times_\R a_j$ is defined as real multiplication, as is appropriate (both $\lambda$ and $a_j$ are real numbers).

With this set of definitions, the structure $\struct {\R^n, +, \cdot}$ is a vector space, as is shown in Proof of Real Vector Space below.

### Proof of Real Vector Space

In order to show that $\struct {\R^n, +, \cdot}$ is a vector space, we need to show that:

$\forall \mathbf x, \mathbf y \in \R^n, \forall \lambda, \mu \in \R$:

$(1): \quad \lambda \cdot \paren {\mathbf x + \mathbf y} = \paren {\lambda \cdot \mathbf x} + \paren {\lambda \cdot \mathbf y}$
$(2): \quad \paren {\lambda +_\R \mu} \cdot x = \paren {\lambda \cdot \mathbf x} + \paren {\mu \cdot \mathbf x}$
$(3): \quad \paren {\lambda \times_\R \mu} \cdot x = \lambda \cdot \paren {\mu \cdot \mathbf x}$
$(4): \quad \forall \mathbf x \in \R^n: 1 \cdot \mathbf x = \mathbf x$.

where $1$ in this context means the real number one.

From External Direct Product of Groups is Group, we have that $\struct {\R^n, +}$ is a group in its own right.

Let:

$\mathbf x = \tuple {x_1, x_2, \ldots, x_n}$
$\mathbf y = \tuple {y_1, y_2, \ldots, y_n}$

Checking the criteria in turn:

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

 $\ds \lambda \cdot \paren {\mathbf x + \mathbf y}$ $=$ $\ds \tuple {\lambda \times_\R \paren {x_1 +_G y_1}, \lambda \times_\R \paren {x_2 +_G y_2}, \ldots, \lambda \times_\R \paren {x_n +_G y_n} }$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \tuple {\paren {\lambda \times_\R x_1 +_G \lambda \times_\R y_1}, \paren {\lambda \times_\R x_2 +_G \lambda \times_\R y_2}, \ldots, \paren {\lambda \times_\R x_n +_G \lambda \times_\R y_n} }$ $\times_\R$ distributes over $+_G$ $\ds$ $=$ $\ds \tuple {\lambda \times_\R x_1, \lambda \times_\R x_2, \ldots, \lambda \times_\R x_n} + \tuple {\lambda \times_\R y_1, \lambda \times_\R y_2, \ldots, \lambda \times_\R y_n}$ Definition of $+$ over $\R^n$ $\ds$ $=$ $\ds \lambda \cdot \tuple {x_1, x_2, \ldots, x_n} + \lambda \cdot \tuple {y_1, y_2, \ldots, y_n}$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \lambda \cdot \mathbf x + \lambda \cdot \mathbf y$ Definition of $\mathbf x$ and $\mathbf y$

So $(1)$ has been shown to hold.

$(2): \quad \paren {\lambda +_\R \mu} \cdot x = \paren {\lambda \cdot \mathbf x} + \paren {\mu \cdot \mathbf x}$:

 $\ds \paren {\lambda +_\R \mu} \cdot x$ $=$ $\ds \tuple {\paren {\lambda +_\R \mu} \times_\R x_1, \paren {\lambda +_\R \mu} \times_\R x_2, \ldots, \paren {\lambda +_\R \mu} \times_\R x_n}$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \tuple {\paren {\lambda \times_\R x_1 +_G \mu \times_\R x_1}, \paren {\lambda \times_\R x_2 +_G \mu \times_\R x_2}, \ldots, \paren {\lambda \times_\R x_n +_G \mu \times_\R x_n} }$ $\times_\R$ distributes over $+_\R$, and $+_\R$ is the same operation as $+_G$ $\ds$ $=$ $\ds \tuple {\lambda \times_\R x_1, \lambda \times_\R x_2, \ldots, \lambda \times_\R x_n} + \tuple {\mu \times_\R x_1, \mu \times_\R x_2, \ldots, \mu \times_\R x_n}$ Definition of $+$ over $\R^n$ $\ds$ $=$ $\ds \lambda \cdot \tuple {x_1, x_2, \ldots, x_n} + \mu \cdot \tuple {x_1, x_2, \ldots, x_n}$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \lambda \cdot \mathbf x + \mu \cdot \mathbf x$ Definition of $\mathbf x$ and $\mathbf y$

So $(2)$ has been shown to hold.

$(3): \quad \paren {\lambda \times_\R \mu} \cdot x = \lambda \cdot \paren {\mu \cdot \mathbf x}$:

 $\ds \paren {\lambda \times_\R \mu} \cdot x$ $=$ $\ds \tuple {\paren {\lambda \times_\R \mu} \times_\R x_1, \paren {\lambda \times_\R \mu} \times_\R x_2, \ldots, \paren {\lambda \times_\R \mu} \times_\R x_n}$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \tuple {\lambda \times_\R \paren {\mu \times_\R x_1}, \lambda \times_\R \paren {\mu \times_\R x_2}, \ldots, \lambda \times_\R \paren {\mu \times_\R x_n} }$ Real Multiplication is Associative $\ds$ $=$ $\ds \lambda \cdot \tuple {\mu \times_\R x_1, \mu \times_\R x_2, \ldots, \mu \times_\R x_n}$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \lambda \cdot \paren {\mu \cdot \tuple {x_1, x_2, \ldots, x_n} }$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \lambda \cdot \paren {\mu \cdot \mathbf x}$ Definition of $\mathbf x$

So $(3)$ has been shown to hold.

$(4): \quad \forall \mathbf x \in \R^n: 1 \cdot \mathbf x = \mathbf x$:

 $\ds 1 \cdot \mathbf x = \mathbf x$ $=$ $\ds \tuple {1 \times_\R x_1, 1 \times_\R x_2, \ldots, 1 \times_\R x_n}$ Definition of $\cdot$ over $\R \times \R^n$ $\ds$ $=$ $\ds \tuple {x_1, x_2, \ldots, x_n}$ Real Multiplication Identity is One $\ds$ $=$ $\ds \mathbf x$ Definition of $\mathbf x$

So $(4)$ has been shown to hold.

So the $\R$-module $\R^n$ is a vector space, as we were to prove.

$\blacksquare$

## Warning

Notice that the factors of $\R^n$ are considered to be elements of a group, where $+_\R$ is the only operator defined.

That is, in this particular context, $\times_G$ is not defined and can not be used.

This stands to reason, as in the context of a vector space, there is no unique and canonical definition for multiplication of vectors.

Several are defined (for example, dot product and cross product), but these are not canonical.

## Also see

It follows directly, by setting $n = 1$, that the $\R$-module $\R$ itself can also be regarded as a vector space.

This is expanded upon in Real Numbers form Vector Space.