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). |
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.
Related post:
- New technique developed to detect autism in children – EurekAlert
- Gujarat Board to offer Mathematics to Non-Science Students from this year onwards – Jagran Josh
- Institute of Mathematics & Application (IMA) Recruitment 2019 for Professor Posts – Jagran Josh
- Help with primary school mathematics — September is just nine weeks away – Galway Advertiser
- Government Launches Effort to Strengthen Math Skills & Improve Job Prospects – Government of Ontario News
- Chaiwalla to Doctor: 5 Teachers Providing Free JEE/NEET Coaching to Needy Students – The Better India
- A celebration of Science, Technology, Engineering, and Mathematics (STEM) – Daily Trust
- Sum of a life – THE WEEK
- Assistant Professor (Tenure Track) in Mathematics in South Holland, Delft – IamExpat in the Netherlands
- Standing in Galileo’s shadow: Why Thomas Harriot should take his place in the scientific hall of fame – OUPblog