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Suppose $\{v_{1}, v_{2}, \ldots, v_{n}\}$ span a vector space $V$. If I know that the dimension of $V$ is $n$, why must the ${v_{i}}$'s form a basis?

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Because any spanning set contains a basis, and every basis has exactly $\dim(V)$ vectors in it. –  Arturo Magidin Oct 13 '11 at 21:15
    
@user1205 The following exercise may also be helpful: Suppose $V$ is a finite dimensional vector space of dimension $n$ and let $\{v_1 \ldots v_n\}$ be $n$ linearly independent vectors in $v$. Why must they form a basis? –  user38268 Oct 13 '11 at 21:54

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If they were linearly dependent (the only thing missing to make them a basis), then one of the vectors is a linear combination of the others. So if this vector is tossed out the remaining vectors still span. You can continue this process until all "redundant" vectors are gone. You'll be left with a linearly independent spanning set (a basis) with less than $n$ vectors! (which is impossible since the dimension -- the size of every basis -- is $n$)

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So you have n linearly independent vectors in an n dimensional vector space. Further, you have that they span the whole space. So everything in the space can be written as a linear combination of these n vectors - hence the n vectors at least contain a basis.

But it is known that $n - 1 $ vectors can never span a space of dimension $n$ or greater. So it must be minimal as well.

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