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I've been asked to show that the dimension of a certain subspace (intersection of 2 subspaces) is actually 1. My understanding is that the dimension is the number of vectors in the basis of a subspace. Given that $ X = \langle v \rangle$ and $v$ is given explicitly, I don't know what more to do to answer the question. How do I show something that seems to require little more than observation?

Or perhaps, I simply do not understand.

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Let me show you the question:

I discovered that given two subspaces $U_1, U_2$ the following was true:

$$ U_1 \cap U_2 = \langle (1, 2, 0, 3) \rangle$$ (a column vector)

And the follow up question says show that $ \dim (U_1 \cap U_2) = 1$ and $ \dim (U_2) = 2$ where $U_2 = \langle (1, 2, 0, 3), (1, 0, 1, 1) \rangle$ (also column vectors). Now to me its pretty obvious that $ \dim (U_1 \cap U_2) = 1$ and $ \dim (U_2) = 2$ are true just by looking at the findings.

Or should I first go about showing that these sets are indeed the basis of their respective subspaces and then state that these are one-/two- element sets and consequently the statements about the dimensions are true or...?

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  • $\begingroup$ Could you ask the question more specifically? What vector in what space? $\endgroup$ – Ian Coley Sep 15 '13 at 19:25
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    $\begingroup$ If you know that $X$ is the span of $\{v\}$, then yes, there’s nothing more to say. However, ‘dimension of a vector’ and ‘intersection of 2 vectors’ don’t really make sense, so perhaps you should say exactly how $X$ was defined. $\endgroup$ – Brian M. Scott Sep 15 '13 at 19:25
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    $\begingroup$ I agree that if you’ve shown that $U_1\cap U_2=\langle(1,2,0,3)\rangle$, then it’s obvious that $\dim(U_1\cap U_2)=1$. In this case it’s also pretty obvious that $\dim U_2=2$, though technically you do have to show that those two vectors are linearly independent. It’s trivial, but some it’s technically necessary. $\endgroup$ – Brian M. Scott Sep 15 '13 at 19:40
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    $\begingroup$ To show that $\dim ⟨(1,2,0,3)⟩ = 1$, maybe you should remark that $(1,2,0,3)$ is not zero. And similarly to show that $\dim ⟨(1,2,0,3),(1,0,1,1)⟩ = 2$ you should remark that these two vectors are linearly independent. $\endgroup$ – GEdgar Sep 15 '13 at 19:43
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    $\begingroup$ @Siyanda: You’re welcome. (‘Sometimes it really is that simple’ is one of my favorite observations, too.) $\endgroup$ – Brian M. Scott Sep 15 '13 at 19:50
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If you’ve shown that $U_1\cap U_2=\langle(1,2,0,3)\rangle$, then it is obvious that $\operatorname{dim}(U_1\cap U_2)=1$; the only additional comment that you might want to make is that $(1,2,0,3)$ isn’t the zero vector, since if it were, the dimension of the subspace would be $0$. Here it’s also pretty obvious that $\operatorname{dim}U_2=2$, though technically you do have to show that those two vectors really are linearly independent. With just two vectors this is especially easy: a set of two vectors is linearly independent if and only if neither vector is a scalar multiple of the other. (If you’ve not seen this fact before, you should try to prove it; it’s not at all hard to do.)

A comment on the terminology in your original question: the dimension of a certain vector doesn’t make sense. Vector spaces have dimensions; individual vectors don’t.

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