# Is $\mathbb{R}^1$ a subspace of $\mathbb{R}^2?$

My intuition tells me it is.

But in terms of vectors, the span of a vector with only one component (a vector in $\mathbb{R}^1$) is not said to be a subspace of $\mathbb{R}^2$

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What is the definition of a linear subspace? –  AnonymousCoward Oct 7 '11 at 1:39
No, $\mathbb{R}$ is not even a subset of $\mathbb{R}^2$. However, you do have subspaces of $\mathbb{R}^2$ that are isomorphic to $\mathbb{R}$, the set $\{[r,0]\mid r\in\mathbb{R}\}$ is one in particular from the isomorphism $r\mapsto [r,0]$. –  yunone Oct 7 '11 at 1:42

Technically, the answer is no, because $\mathbb{R}^1$ is not a subset of $\mathbb{R}^2$. $\mathbb{R}^1$ consists of real numbers, whereas $\mathbb{R}^2$ consists of ordered pairs of real numbers; therefore $\mathbb{R}^1$ is not contained in $\mathbb{R}^2$.

However, there are many ways one can "put a copy of $\mathbb{R}^1$ into $\mathbb{R}^2$", and depending on how this is done, the result may or may not be a subspace of $\mathbb{R}^2$. Look at the definition of a subspace, and consider that $$S=\{(x,0)\in\mathbb{R}^2\mid x\in\mathbb{R}^1\}$$ is a subspace of $\mathbb{R}^2$ because

• the zero vector, $(0,0)\in S$
• for any two $(a,0),(b,0)\in S$, we have that $(a,0)+(b,0)=(a+b,0)\in S$
• for any $(a,0)\in S$ and $c\in\mathbb{R}$, we have that $c(a,0)=(ca,0)\in S$

while

$$T=\{(x,1)\in\mathbb{R}^2\mid x\in\mathbb{R}^1\}$$

completely fails to be a subspace of $\mathbb{R}^2$ - every condition is false:

• the zero vector, $(0,0)\notin T$
• for any two $(a,1),(b,1)\in T$, we don't have that $(a,1)+(b,1)=(a+b,2)\in T$
• for any $(a,1)\in T$ and $c\in\mathbb{R}$, we don't have that $c(a,1)=(ca,c)\in T$ (unless $c=1$)

Generally, you should note that it is entirely possible for a subspace of a vector space $V$ to have smaller dimension than $V$; for example, for any vector space $V$, the subset consisting of only the zero element is always a subspace, and it has dimension 0.

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Haha, thanks for the edit, Henning :) –  Zev Chonoles Oct 7 '11 at 1:57
hmm, ok that makes a lot of sense. Thanks so much! –  weezybizzle Oct 7 '11 at 2:09
You ain't getting a better answer then that. Good job,guys! –  Mathemagician1234 Oct 7 '11 at 4:48
Not to mention that there are maps $\mathbb R\to\mathbb R^2$ whose image is not even an affine subspace. –  Rasmus Oct 7 '11 at 7:19