# The Vector Space of Lines Through The Origin of R^{2}

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 that the set of lines that passes through the origin in $\mathbb{R}^2$ form a vector space. Let $L$ denote this set of lines, and let $u, v, w \in L$ such that $u = mx + ny$, $v = px + qy$ and $w = rx + sy$, and let $a, b \in \mathbb{F}$. Define addition by standard addition and scalar multiplication by standard scalar multiplication.

**1.**$u + v = (mx + ny) + (px + qy) = (m + p)x + (n + q)y = (p + m)x + (q + n)y = (px + qy) + (mx + ny) = v + u$.

**2.**$u + (v + w) = (mx + ny) + (p + r)x + (q + s)y = (m + [p + r])x + (n + [q + s])y$- $= ([m + p] + r)x + ([n + q] + s)y = [m + p]x + [n + q]y + (rx + sy) = (u + v) + w$.

**3.**The zero line vector is $0 = 0x + 0y$, and this $u + 0 = (m + 0)x + (n + 0)y = mx + my = u$.

**4.**The additive inverse of $u$ is $-u = -mx -ny$. That is $u + (-u) = (m -m)x + (n - n)y = 0x + 0y = 0$.

**5.**$a(bu) = a(bmx + bny) = abmx + abny = (ab)mx + (ab)my = (ab)u$.

**6.**The multiplicative identity is the scalar $k = 1$, that is $1u = 1(mx + ny) = mx + ny = u$.

**7.**$a(u + v) = a[(mx + ny) + (px + qy)] = a(mx + ny) + a(px + qy) = au + av$.

**8.**$(a + b)u = (a + b)(mx + ny) = a(mx + ny) + b(mx + ny) = au + bu$.

**9.**$u + v = (m + p)x + (n + q)y$ is a line in $\mathbb{R}^2$ and so $(u + v) \in L$.

**10.**$au = (am)x + (an)y$ is a line in $\mathbb{R}^2$ and so $(au) \in L$.

Since the set $L$ of lines in $\mathbb{R}^2$ satisfies all ten vector space axioms under the defined operations of addition and multiplication, we have that thus $L$ is a vector space.

# The Vector Space of Planes Through The Origin of R^{3}

Let $\Pi$ be the set of all planes that pass through through the origin in $\mathbb{R}^3$. It can also be shown that $\Pi$ forms a vector space. We will omit verifying all ten axioms since they're similar to verifying the axioms that show the set of lines $L$ in $\mathbb{R}^2$ form a vector space.