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:

$(1): \quad \left({\R^2, +, \times}\right)_\R$

$(2): \quad \left\{{0}\right\}$

$(3): \quad$ A line through the origin.

Proof 1

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, then:

$\left({\gamma_1, \gamma_2}\right) = \lambda \times \left({\alpha_1, \alpha_2}\right) + \mu \times \left({\beta_1, \beta_2}\right)$

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

which we get by solving the simultaneous eqns:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \alpha_1 \times \lambda + \beta_1 \times \mu\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \alpha_2 \times \lambda + \beta_2 \times \mu\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

The result follows.

$\blacksquare$

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:

$(1): \quad \left({\R^2, +, \times}\right)_\R$

$(2): \quad \left\{{0}\right\}$

$(3): \quad$ A line through the origin.

Proof

<onlyinclude> Follows directly from Dimension of Proper Subspace Less Than its Superspace.

$\blacksquare$ <onlyinclude>