Subspaces of Dimension 2 Real Vector Space/Proof 1

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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

Let $S$ be a non-zero subspace of $\struct {\R^2, +, \times}_\R$.

Then $S$ contains a non-zero vector $\tuple {\alpha_1, \alpha_2}$.

Hence $S$ also contains $\set {\lambda \times \tuple {\alpha_1, \alpha_2}, \lambda \in \R}$.

From Equation of Straight Line in Plane, this set may be described as a line through the origin.


Suppose $S$ also contains a non-zero vector $\tuple {\beta_1, \beta_2}$ which is not on that line.

Then:

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

Otherwise $\tuple {\beta_1, \beta_2}$ would be $\zeta \times \tuple {\alpha_1, \alpha_2}$, 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 $S = \struct {\R^2, +, \times}_\R$.

Because, if $\tuple {\gamma_1, \gamma_2}$ is any vector at all, then:

$\tuple {\gamma_1, \gamma_2} = \lambda \times \tuple {\alpha_1, \alpha_2} + \mu \times \tuple {\beta_1, \beta_2}$

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 equations:

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

The result follows.

$\blacksquare$


Sources

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