If $V$ has a basis of size $n$, show that any linearly independent set of size $n$ in $V$ is a basis of $V$. If $V$ has a basis of size $n$, show that any linearly independent set of size $n$ in $V$ is a basis of $V$.
I can see this is true, but don't know how to prove it, any hints?
Thanks
 A: Suppose you've a set with $\;n\;$ linearly independent vectors $\;B:=\{v_1,...,v_n\}\;$ , If $\;Span(B)\neq V\;$, there exists $\;v\in V\;$ s.t. $\;v\notin Span(B)\;$ . But this last is equivalent to $\;B\cup\{v\}\;$ linearly independent, which is impossible in a linear space of dimension $\;n\;$ ... Fill in details now.
A: There is nothing to prove since the dimension of $V$ is $n$ and the fact that any set of $n$ linearly independent vectors form a basis is immediately true by the definition of basis.
Notably note that any set of $n$ linearly independent vectors span $V$, indeed since $V$ has a basis of $n$ vectors if the new set didn't span $V$ we would have a contradiction.
A: I disagree with gimusi that it follows from the definition of "basis".  The definition of "basis" that he links to says that a basis is a set of vectors that (1) spans the space and (2) are independent.  However, it does follow from the definition of "dimension"!  It can be shown that all bases for a given vector space have the same number of members and we call that the "dimension" of the vector space.  Basically, a basis for a vector space of dimension n has three properties: (1) they span the space, (2) they are independent, and (3) there are n vectors in the set.  And if any two of those are true then the third is true!
