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In my attempts to learn linear algebra from Khan Academy I've come across several concepts that I can't completely connect.

Every vector space has a base. This base consists of the minimal elements that can span the entire vector space.

The number of elements in this vector space is defined to be the vector space's dimension.

When trying to conclude the power of a vector space (as in how many vectors there are in the space), I know that $|V| = |F| ^ n$

I also know that this small n = dimension of V. What I don't understand is why does the dimension equals that $n$?

Thanks!

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You are talking about two different things. The dimension of a vector space V is the number of elements of its basis (or rather of any basis), not the number of elements in V. –  user127.0.0.1 Dec 4 '12 at 17:08
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The subject is a bit confused. The dimension is not "$\mathbb R^n$. You are confusing terminology there –  Thomas Andrews Dec 4 '12 at 17:12
    
I know, what I said is that dimension equals N, the power of R. Maybe the question isn't clear enough. I'll edit it –  vondip Dec 4 '12 at 17:15
    
You are right, I just edited my question. thanks! –  vondip Dec 4 '12 at 17:18
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what do you mean ? –  vondip Dec 4 '12 at 17:35

1 Answer 1

up vote 3 down vote accepted

A basis for a vector space is a linearly independent set that spans the space. For the Euclidean space $\mathbb R^n$, a basis is $(1,0,...,0),(0,1,0,...,0),...,(0,...,0,1)$. Since there are $n$ elements in this set, the dimension of $\mathbb R^n$ is $n$.

Bases for a vector space are not unique. It can be shown that any two bases have the same number of elements, so that this dimension is well-defined.

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while this makes sense, if I were to say that I have a vector space which follows a certain rule such that it only has 2 vectors in its base, why would that mean that the entire space is only F^2? Why wouldn't there be more? –  vondip Dec 4 '12 at 17:48
    
Is there a particular name for this basis in your answer? Something like 'canonical vector space basis'? –  000 Dec 4 '12 at 17:54

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