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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 $R^2$?

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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
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His linear algebra prof....or any other mathematician, of course. –  DonAntonio Dec 3 '12 at 12:56

4 Answers 4

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.

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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.

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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 $\mathbf C$.

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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.

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

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