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What is the difference (or connection) between the dimension of a vector space and the dimension in terms of bases?

For instance, when we talk about the vector space $\mathbb{R}^3$, we are talking about a 3-dimensional vector space. This vector space contains vectors with three elements: $(x_1, x_2, x_3)$.

But we also know that the dimension of a vector space is equal to the number of vectors in its basis. For instance, let's say we have the basis $\{(1, 1, -2, 0, -1), (0, 1, 2, -4, 2), (0, 0, 1, 1, 0) \}$. These vectors contain 5 elements $(x_1, x_2, x_3, x_4, x_5)$, so the vector space is 5-dimensional or $\mathbb{R}^5$. But, by the definition of dimension in terms of bases, this vector space has a dimension of 3 since the basis contains 3 vectors; so the vector space should be $\mathbb{R}^3$?

I hope that I effectively conveyed my confusion. I would greatly appreciate it if people could please take the time to clarify my misunderstanding and elaborate on the differences and/or connections between these concepts.

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    The dimension of a vector space is defined over the number of elements of the basis. Here, doesn't matter the number of cordinates in the vectors. In your examples, the basis that you write is a basis of a subspace of $\mathbb{R}^5$ such that have dimension 3. – 2017-02-22

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Consider the vectors $e_1 =(1,0,0)$, $e_2 = (0,1,0)$ and $e_3 = (0,0,1)$ in $\mathbf{R}^3$. Then the vectors $e_1, e_2, e_3$ form a basis for the whole space, which has dimension $3$. This case is not a problem for you.

But what about the vectors $e_1$ and $e_2$? This is more like the case that is not clear to you. In this case, the vectors $e_1$ and $e_2$ do not form a basis for the whole space. Instead, they form a basis for the plane with equation $z = 0$. In other words, they form a basis for a certain subspace of the whole space. Since this subspace is a plane, it should be unsurprising that it has dimension $2$.

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    So, in my example, the vectors $\{(1, 1, -2, 0, -1), (0, 1, 2, -4, 2), (0, 0, 1, 1, 0) \}$ are in a 5-dimensional vector space $\mathbb{R}^5$, but they are a basis for a 3-dimensional subspace $\mathbb{R}^3$? – 2017-02-22
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    I haven't done the computations to check that the three vectors are linearly independent. But if they are, then they form a basis for a three-dimensional subspace of $\mathbf{R}^5$. This three-dimensional space consists of *certain quintuples* $(x,y,z,u,v)$ of real numbers. It is not $\mathbf{R}^3$, which is the set of *all triples* of real numbers $(x,y,z)$. – 2017-02-22
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    Ok, I understand. In your example, the vector space itself is 5-dimensonal, but the subspace is 3-dimensional? – 2017-02-22
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    Well, yes, but it's your example. – 2017-02-22
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    Ok, I understand. Thank you for the assistance. – 2017-02-22
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Every $\mathbb R^n$ is a vector space (or at least can be a vector space), but not every vector space is $\mathbb R^n.$

In your example with the three vectors $\{(1, 1, -2, 0, -1), (0, 1, 2, -4, 2), (0, 0, 1, 1, 0) \}$, the vectors belong to $\mathbb R^5,$ which we can consider to be a $5$-dimensional vector space, but the span of the three vectors also is a vector space, because it satisfies all of the criteria for the definition of a vector space.

Since these three vectors are independent, they form a basis of a $3$-dimensional vector space. Yet that vector space is most definitely not $\mathbb R^3,$ since $(1, 1, -2, 0, -1)$, for example, is not in $\mathbb R^3.$

In fact there are infinitely many $3$-dimensional vector spaces, and only one of those is $\mathbb R^3.$ The only reason you hear so much about $\mathbb R^3$ is that we are so terribly prejudiced in its favor. (It is relatively simple to work with, so there's some reason for the prejudice.)

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    Your post is illuminating. Thank you. – 2017-02-22
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The vector space $\Bbb R^n$ means exactly one thing: $n$-tuples of numbers with entries in $\Bbb R$. (This can be formalized by saying an element of $\Bbb R^n$ is really a function from $\{1, 2, \ldots, n\}$ to $\Bbb R$: For example the vector $(\sqrt{2}, -3, \pi)$ is really the function $1 \mapsto \sqrt{2},\, 2\mapsto -3,\, 3 \mapsto \pi$) although this isn't necessarily something you should worry about.)

Every $d$-dimensional subspace of $\Bbb R^n$ is isomorphic to ("functionally equivalent to") $\Bbb R^d$, although if $n \neq d$, it is not literally "equal to" $\Bbb R^d$: it doesn't consist of $d$-tuples of real numbers, but instead $n$-tuples.