The Vector Space of n-Component Vectors

# The Vector Space of n-Component Vectors

From the Vector Spaces page, recall the definition of a Vector Space:

 Definition: A nonempty set $V$ is considered a vector space if the two operations: 1. addition of the objects $\mathbf{u}$ and $\mathbf{v}$ that produces the sum $\mathbf{u} + \mathbf{v}$, and, 2. multiplication of these objects $\mathbf{u}$ with a scalar $a$ that produces the product $a \mathbf{u}$, are both defined and the ten axioms below hold. Furthermore, if $V$ is a vector space then the objects in $V$ are called vectors: 1. $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ (Commutativity of vector addition). 2. $\mathbf{u} + (\mathbf{v} + \mathbf{w}) = (\mathbf{u} + \mathbf{v}) + \mathbf{w}$ (Associativity of vector addition). 3. There exists a zero vector $\mathbf{0}$ such that $\mathbf{0} + \mathbf{u} = \mathbf{u} + \mathbf{0} = \mathbf{u}$ (Existence of an additive identity). 4. For every $\mathbf{u} \in V$, there exists a vector $-\mathbf{u}$ such that $\mathbf{u} + (-\mathbf{u}) = (-\mathbf{u}) + \mathbf{u} = \mathbf{0}$ (Existence of an additive inverses). 5. $a(b\mathbf{u}) = (ab)\mathbf{u}$. (Associativity of scalar multiplication) 6. $1\mathbf{u} = \mathbf{u}$ (Existence of a multiplicative identity). 7. $a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}$ (Distributivity of a scalar multiplication over vector addition). 8. $(a + b)\mathbf{u} = a\mathbf{u} + b\mathbf{u}$. (Distributivity of scalar multiplication over field addition) 9. If $\mathbf{u}, \mathbf{v} \in V$, then $(\mathbf{u} + \mathbf{v}) \in V$ (Closure under addition). 10. If $a$ is any scalar and $\mathbf{u} \in V$, then $a\mathbf{u} \in V$ (Closure under scalar multiplication).

We will now verify the set of n-component vectors or "lists" as a vector space.

Let $\mathbb{F}^n$ be the set of n-component vectors whose components come from either the set of real numbers or complex numbers. Let $u, v, w \in \mathbb{F}^n$ such that $u = (u_1, u_2, ..., u_n)$, $v = (v_1, v_2, ..., v_n$ and $w = (w_1, w_2, ..., w_n)$, and define addition to be standard vector addition, and scalar multiplication to be standard scalar multiplication. Let $a, b \in \mathbb{F}$ be scalars from the field of real numbers of complex numbers.

We will now verify all ten axioms to show that $\mathbb{F}^n$ is a vector space.

• 1. $u + v = (u_1 + v_1, u_2 + v_2, ..., u_n + v_n) = (v_1 + u_1, v_2 + u_2, ..., v_n + u_n) = v + u$.
• 2. $u + (v + w) = (u_1 + [v_1 + w_1], u_2 + [v_2 + w_2], ..., u_n + [v_n + w_n]) = ([u_1 + v_1] + w_1, [u_2 + v_2] + w_2, ..., [u_n + v_n] + w_n) = (u + v) + w$.
• 3. Let $0 = \underbrace{(0, 0, ..., 0)}_{\mathrm{n-times}}$. Then $u + 0 = (u_1, u_2, ..., u_n) + (0, 0, ..., 0) = (u_1 + 0, u_2 + 0, ..., u_n + 0) = u$.
• 4. The additive inverse of $u$ is $-u = (-u_1, -u_2, ..., -u_n)$ and so $u + (-u) = (u_1 - u_1, u_2 - u_2, ..., u_n - u_n) = (0, 0, ..., 0) = 0$.
• 5. $k(lu) = a(bu_1, bu_2, ..., bu_n) = (abu_1, abu_2, ..., abu_n) = ([ab]u_1, [ab]u_2, ..., [ab]u_n) = (ab)(u_1, u_2, ..., u_n) = (ab)u$.
• 6. The scalar $1$ is a multiplicative identity, that is $1u = 1(u_1, u_2, ..., u_n) = u$.
• 7. $a(u + v) = a(u_1 + v_1, u_2 + v_2, ..., u_n + v_n) = (au_1 + av_1, au_2 + av_2, ..., au_n + av_n) = (au_1, au_2, ..., au_n) + (av_1, av_2, ..., av_n) = au + av$.
• 8.
(1)
\begin{align} \quad (a+b)u = ([a+b]u_1, [a+b]u_2, ..., [a+b]u_n) = (au_1 + bu_1, au_2 + bu_2, ..., au_n + bu_n) \\ = (au_1, au_2, ..., au_n) + (bu_1, bu_2, ..., bu_n) = au + bu \end{align}
• 9. Since $u + v = (u_1 + v_1, u_2 + v_2, ..., u_n + v_n)$, we have that $(u + v) \in \mathbb{F}^n$.
• 10. Since $au = (au_1, au_2, ..., au_n)$ we have that $(au) \in \mathbb{F}^n$.

Since $\mathbb{F}^n$ under the defined operations of addition and multiplication satisfies all ten axioms of a vector space, then we have that $\mathbb{F}^n$ is in fact a vector space.