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Let's say I have the following family

$f = ((1,0,1,0),(-1,0,-1,0), (1,1,1,1),(0,1,0,1), (1,2,1,0))$.

I want to first find its rank. I can say, by eliminating all the colinear vectors, that the rank of my family is equal to

rank of $f = \dim(span((1,0,1,0),(0,1,0,1), (1,2,1,0))) = 3$.

Now, I'm very confused about the the whole idea of dimensions. When someone say that the dimension of a basis is always equal to the dimension of the vector space, they're speaking about the number of vectors in the basis, correct? Does this mean that a family of $3$ vectors can be a basis, even if the vectors in the family are of dimension $4$.

In the particular example I gave, can I say that $((1,0,1,0),(0,1,0,1),(1,2,1,0))$ is a basis of $\mathbb R$$^3$? Or does that make no sense? Thank you.

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  • $\begingroup$ I think you are confusing basis of a vector space with basis of a subspace. One has dimension $4$, the other might have only dimension $3$. $\endgroup$ – Dietrich Burde Apr 2 '18 at 18:52
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The dimension of a finite-dimensional vector space is the number of vectors in a basis for the vector space. This number is independent of the choice of the basis.

You can't say that "$\{(1,0,1,0),(0,1,0,1),(1,2,1,0)\}$ is a basis of $\Bbb{R}^3$" because basis vectors are "representatives" of the vector space, so they must live in that vector space. However, in this numerical example, the proposed vectors live in $\Bbb{R}^4$. You may rather say that this set of linear independent vectors spans a three-dimensional subspace of $\Bbb{R}^4$.

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  • $\begingroup$ Thanks for your response. Ok, but the exercise I have asks to find the rank of the family (which I've done) and then give a base of the vector space it creates. I'm not exactly sure what that's supposed to mean, could you point me towards the right direction? Thanks. $\endgroup$ – iaskdumbstuff Apr 2 '18 at 19:03
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    $\begingroup$ @user494405 Your proposed basis is already the answer. It's clearly linearly independent. My point is that it is a basis for the range of $f$. (not $\Bbb{R}^3$) $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 2 '18 at 19:07
  • $\begingroup$ Sorry if I'm being dense, but I don't exactly understand what it means to be a basis for a range. I thought the range was just a number, like dimension? $\endgroup$ – iaskdumbstuff Apr 2 '18 at 19:12
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    $\begingroup$ @user494405 Oh sorry, I mistakenly thought that $f$ is a function. Then my point should be "it is a basis for $\mathop{\rm span}(f)$". Of course, the range of a function is the image of its domain, thus a subset of its codomain. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 2 '18 at 19:15
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    $\begingroup$ Ohhhh, I think I understand! Doesn't it seem a little superfluous to ask to find the rank of a family and a basis of the vector space it creates (which is span(f), right?), because to find the rank, I need to find the span, which in turn gives me the basis of the span(f)? I'm not sure if that made sense or not, but either way, thank you very much for the help! $\endgroup$ – iaskdumbstuff Apr 2 '18 at 19:21
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$\mathbb R$$^3$ has dimension 3, but that doesn't mean that a space with dimension 3 is $\mathbb R$$^3$. All vector spaces of finite dimension are isomorphic to all vector spaces of the same dimension (given a fixed scalar field), but they aren't equal. While each finite dimension has a canonical vector space, and we can speak of "the" vector space of that dimension, we can do so only up to isomorphism. So ((1,0,1,0),(0,1,0,1),(1,2,1,0)) is the basis of a space that is isomorphic to $\mathbb R$$^3$, but since the vectors aren't in $\mathbb R$$^3$, the space is not equal to $\mathbb R$$^3$. There are multiple 3 dimensional subspaces of $\mathbb R$$^4$, and they are all isomorphic to each other, but they are not equal to each other.

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