How to Show that the Only Subspaces of $R^2$ are the zero subspace, $R^2$ itself, and the lines through the origin

Solution 1:

Let me give a more algebraic-flavor proof (without regarding too much about the geometry).

Let $S$ be a subspace of $\mathbb{R}^2$, by definition, $(0, 0) \in S$.

If $S$ contains only one element, then $S = \{(0, 0)\}$, i.e., the zero subspace.

Now suppose that $S$ contains more than one element, i.e., there exists $u = (x_0, y_0) \in S$ and $u \neq 0$. Then since $S$ is a subspace, for any $\lambda \in \mathbb{R}$, $\lambda u \in S$, therefore $S$ contains the line connects the origin and $u$, call it $L$.

If $L \subsetneqq S$, i.e., $S$ contains another point $v = (x_1, y_1)$ which is not a multiple of $u$. We shall show under this case $S = \mathbb{R}^2$. To show this, we need to show for any $(a, b) \in \mathbb{R}^2$, $(a, b) \in S$. It can be shown that there exist $\lambda \in \mathbb{R}$ and $\mu \in \mathbb{R}$ such that $(a, b) = \lambda u + \mu v$. To verify this assertion, it is easy to see that aforementioned $\lambda$ and $\mu$ can be solved from the system $$\begin{cases} a = \lambda x_0 + \mu x_1 \\ b = \lambda y_0 + \mu y_1 \\ \end{cases}$$

Since $u$ and $v$ are not linearly dependent, the above system has a unique solution. Therefore, $(a, b)$ is a linear combination of $u$ and $v$, thus belongs to $S$ for $S$ is a subspace.

The proof is complete.

Solution 2:

Hint: Basically your idea is right. Let me say it this way instead: Suppose it is not zero. Then it contains a vector $v$. So it is at least a line. If it contains some other vector not on the line spanned by $v$, then it is all of $R^2$.