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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?

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  • $\begingroup$ 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 $\endgroup$ – Calvin Khor Feb 22 at 15:48
  • $\begingroup$ ok...interesting...so we could add subspace of any of the above vector spaces where the vector space properties hold.... $\endgroup$ – John Proxer Feb 22 at 15:52
  • $\begingroup$ line equation including origin, and plane equation including origin as subspace of 2D coordinate space. $\endgroup$ – John Proxer Feb 22 at 16:03
  • $\begingroup$ Those are all kernels of linear maps $\mathbb R^n \to \mathbb R^m$. $\endgroup$ – Calvin Khor Feb 22 at 16:07
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I think its simpler if you just think about x-2y=0 as a 2-d vector. For example, you can start with (1,2) and then subsequently take the span of this vector. This will give you a vector space (which goes through the origin) which clearly defines your equation. In general, linear algebra is really flexible so you can get really creative with how you define vector spaces.

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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.

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  • $\begingroup$ your first charaterisation doesn't include the line $x=0$ $\endgroup$ – Calvin Khor Feb 22 at 16:47
  • $\begingroup$ Dang it. You're right. $\endgroup$ – Randall Feb 22 at 16:49
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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 $\{k(ax^2+bx+c)\}$.

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