Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we have two vectors in $R^3$, $v=(1,2,0)$ and $u=(5,3,0)$, and if we draw these vectors in $R^3$ they will be in the $xy$-plane or $R^2$ and also these two vectors are not multiples of each other then why can’t we say that these vectors span $R^2$?

share|cite|improve this question
Who says that we "can’t say that"? – draks ... Dec 3 '12 at 12:43
Probably his linear algebra professor :) – Neal Dec 3 '12 at 12:45
His linear algebra prof....or any other mathematician, of course. – DonAntonio Dec 3 '12 at 12:56

Because $\,\Bbb R^3\rlap{\;\;/}\subset \Bbb R^2\,$ , so vectors in the former are not even vectors in the latter.

Note that you cannot draw the given vectors in the plane $\,\Bbb R^2\,$: what you can do is draw their projections on some plane in $\,\Bbb R^3\,$ and identify this plane with $\,\Bbb R^2\,$, but this can be done in an infinite number of different ways.

share|cite|improve this answer

But these vectors span a $2$-dimensional subset (=a plane) of $\mathbb R^3$: Assume that $av + bu = 0$. Then $a + 5b = 0$ and $2a + 3b = 0$. Then $a = -5b$ and hence $2a +3b = -10b + 3b = -7b = 0$ so that $b = 0$ and hence $a=0$. Hence $u$ and $v$ are linearly independent.

share|cite|improve this answer

Those vectors span the image of the most obvious embedding of $\def\R{\mathbf R}\R^2$ in $\R^3$, but since this embedding does not have many properties to make it stand out among other linear embeddings, it is unusual to identify $\R^2$ with a part of $\R^3$ in this way, in the same manner as one identifies for instance $\R$ with a part of $\mathbf C$.

share|cite|improve this answer

They will span a subspace of $\Bbb R^3$ which will look like $\Bbb R^2$. Your intuition is that

$$f:\Bbb R^3 \to \Bbb R^2/f(x,y,z)=(x,y)$$

will be onto, which is the case. But note we don't get an isomorphism for this isn't an injection. You're identifying a plane in $\Bbb R^3$ with $\Bbb R^2$ by a projection, in some sense.

share|cite|improve this answer
Shortly speaking, $\mathbb R^3\cap \mathbb R^2=\emptyset$. – yo' Dec 4 '12 at 8:47
@tohecz What is your point? – Pedro Tamaroff Dec 4 '12 at 14:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.