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

I'm having trouble with a question from an introductory Linear Algebra book. It goes: "Show that the Only Subspaces of $R^2$ are the zero subspace, $R^2$ itself, and the lines through the origin." I'm thinking the easiest way to do this is to show that if $W$ is a subspace of $R^2$ containing $2$ different lines through the origin then $W$ is all of $R^2$. Is this a good way to go about it? and how could I show this?

• The best answer to this question will depend on how much theory you already have available to use. To begin with, what is the definition of a line for you? Perhaps you could even say what book you're using. Commented Jan 26, 2016 at 4:44
• Any subspace of $\Bbb R^2$ has dimension $0,1$, or $2$. Show that 1D subspaces are lines through the origin (almost by definition), and that a 2D subspace must be $\Bbb R^2$ itself. Commented Jan 26, 2016 at 4:46
• Which is this book? Commented Oct 28, 2017 at 6:37
• @ReeshabhRanjan A Course in Linear Algebra by Damiano and Little. Commented Oct 28, 2017 at 23:19

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.

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 $R^2$.

I have the same problem to solve and came up with the following proof. Could anyone please comment on its validity?

Every subspace of $\mathbb{R}^n$ can be described in terms of a span of a nonempty set of vectors $S = \{v_1, v_2, ..., v_k\}$.

Side note: Let $V$ be a subspace of $\mathbb{R}^n$ which can be described in terms of span$(v_1, v_2,...v_k)$. And let $W$ be a second subspace which contains an additional vector $x$ which is not in span$(v_1, v_2,...v_k)$. Then span$(v_1, v_2,...v_k,x)$ describes the subspace $W$.

Now the following cases can arise

1. The set of vectors $S$ contains one element which is $0$. Then the corresponding subspace is the trivial subspace.
2. S contains one vector which is not $0$. In this case the corresponding subspace is a line through the origin.
3. S contains multiple colinear vectors. Same result as 2.
4. S contains multiple vectors of which two form a linearly independent subset. The corresponding subspace is $\mathbb{R}^2$ itself.

Since there are no other possibilities (i.e., there is no linearly independent set of more than 2 vectors in $\mathbb{R}^2$), the proof is complete.