If $V$ has a basis of size $n$, show that any linearly independent set of size $n$ in $V$ is a basis of $V$.

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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!