Subspaces of Dimension 2 Real Vector Space

Theorem
Take the $\mathbb{R}$-vector space $$\left({\mathbb{R}^2, +: \times}\right)_\mathbb{R}$$.

Let $$S$$ be a subspace of $$\left({\mathbb{R}^2, +: \times}\right)_\mathbb{R}$$.

Then $$S$$ is one of:
 * 1) $$\left({\mathbb{R}^2, +: \times}\right)_\mathbb{R}$$;
 * 2) $$\left\{{0}\right\}$$;
 * 3) A line through the origin.

Proof

 * Let $$S$$ be a non-zero subspace of $$\left({\mathbb{R}^2, +: \times}\right)_\mathbb{R}$$.

Then $$S$$ contains a non-zero vector $$\left({\alpha_1, \alpha_2}\right)$$.

Hence $$S$$ also contains $$\left\{{\lambda \times \left({\alpha_1, \alpha_2}\right), \lambda \in \reals}\right\}$$.

This set may be described as a line through the origin.


 * Suppose $$S$$ also contains a non-zero vector $$\left({\beta_1, \beta_2}\right)$$ which is not on that line.

Then $$\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1 \ne 0$$.

Otherwise $$\left({\beta_1, \beta_2}\right)$$ would be $$\zeta \times \left({\alpha_1, \alpha_2}\right)$$, where either $$\zeta = \beta_1 / \alpha_1$$ or $$\zeta = \beta_2 / \alpha_2$$ according to whether $$\alpha_1 \ne 0$$ or $$\alpha_2 \ne 0$$.

But then $$S = \left({\mathbb{R}^2, +: \times}\right)_\mathbb{R}$$.

Because, if $$\left({\gamma_1, \gamma_2}\right)$$ is any vector at all, $$\left({\gamma_1, \gamma_2}\right) = \lambda \times \left({\alpha_1, \alpha_2}\right) + \mu \times \left({\beta_1, \beta_2}\right)$$

where $$\lambda = \frac {\gamma_1 \times \beta_2 - \gamma_2 \times \beta_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}, \mu = \frac {\alpha_1 \times \gamma_2 - \alpha_2 \times \gamma_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}$$

which we get by solving the simultaneous equations:

$$ $$

The result follows.

Alternative Proof
We can also prove this by using the result Dimension of Proper Subspace Less Than its Superspace.