Subspaces of Dimension 2 Real Vector Space

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

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

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

Proof

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

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

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

From Equation of a Straight Line, 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 $$S = \left({\R^2, +, \times}\right)_\R$$.

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.