# What type of Vector Space is equation for a line including the origin?

The equation for a line including the origin is what type of vector space?

I'm told there are a least a few classification of Vector spaces (V):

1. Zero Vector Space ($$\{0\}$$) $$V = \{0\}$$

2. Field Space ($$F$$) $$V = \{all\ \mathbb{R}\}$$ $$V = \{all\ \mathbb{C}\}$$

3. Coordinate Space ($$F^n$$) $$V = \{all\ \mathbb{R}^{n}\}$$ $$V = \{all\ \mathbb{C}^{n}\}$$

4. Matrix Space ($$F^{mxn}$$)

5. Polynomial Space ($$F[x_1, x_2, …, x_n]$$)

6. Function Space (however you denote that)

Which one does a line equation fit into? or is that an as yet unnamed vector space?

I'm thinking its a polynomial space. but, then the book gives the example of a polynomial without an equal sign as being a polynomial vector space.

$$ax^2 + bx + c$$

Which doesn't fit the line equation because it has an equality:

$$x-2y=0$$

so would the line be classified as polynomial space? or something else? or maybe a function space of only one function is a better classification for a line equation?

• A line $ax + by = 0$ is a 1D vector subspace of $\mathbb R^2$. (It is the kernel of the map given by the $1\times 2$ matrix $[a\ \ b]$) There are two special cases, the vertical line $x=0$ and the horizontal line $y=0$ that look almost like $\mathbb R^1$, but are not quite. I would not call your list a classification, but a collection of examples – Calvin Khor Feb 22 at 15:48
• ok...interesting...so we could add subspace of any of the above vector spaces where the vector space properties hold.... – John Proxer Feb 22 at 15:52
• line equation including origin, and plane equation including origin as subspace of 2D coordinate space. – John Proxer Feb 22 at 16:03
• Those are all kernels of linear maps $\mathbb R^n \to \mathbb R^m$. – Calvin Khor Feb 22 at 16:07

A line can be a vector subspace only when it passes through the origin. So, if your line isn't vertical, it must take the form $$y=mx$$ for some $$m$$ (if we're in the plane). Hence $$\{(x,y) \mid y=mx\}$$ should work, and you can check that such a set does in fact form a subspace. If your line were vertical ($$x=0$$), it's simpler: $$\{(0,y) \mid y \in \mathbb{R}\}.$$ Alternatively, you can choose any non-zero vector $$\mathbf{v}$$ on your line and define $$\{c\mathbf{v} \mid c \in \mathbb{R}\}$$ and you should do the same exercise. This version works regardless of the ambient space.
• your first charaterisation doesn't include the line $x=0$ – Calvin Khor Feb 22 at 16:47
A line through the origin can be defined in any vector space - it is simply the set $$\{k\underline{v}\}$$ of the scalar multiples of a given vector $$\{\underline{v}\}$$. A general line (not necessarily through the origin) can also be defined as the set $$\{k\underline{u}+(1-k)\underline{v}\}$$. In a sense, a vector space is a set to which we have added sufficient structure (vector addition and vector multiplication) to define objects that match our intuitive understanding of "lines".
In a vector space of polynomials, each polynomial is a separate vector in its own right. So the line through $$ax^2+bx+c$$ that includes the origin (a.k.a. the zero polynomial) is the set of polynomials $$\{k(ax^2+bx+c)\}$$.