# Question about finite dimensional vector spaces and their bases

Suppose $V$ is a finite dimensional vector space with $\dim V$ = $n$.

Show that:

(i) If $v_{1},\ldots,v_{m} \in V$ are independent, then $m \leq n$ and if $m = n$ they form a basis.

I think I know why this is but I'm not sure if I'm translating it onto paper very well. If $v_{1},\ldots,v_{m} \in V$ are independent, then $m \leq n$ by definition surely? And if $m =n$ then they must form a basis as they are all linearly independent. Do I need to add more?

(ii) If $v_{1},\ldots,v_{m}$ span $V$, then $m \geq n$ and if $m = n$, they form a basis.

I think we explored a theorem that stated that any set of vectors that span $V$ can be reduced to form a basis of $V$, by discarding the necessary vectors - the linearly independent vectors in the set, for example. Furthermore, $m \geq n$ is possible as the spanning set is a linear combination of the linearly independent vectors in the basis.

I think I understand the general idea but I'm not sure if my answers are rigorous or detailed enough, so any help would be much appreciated.

I assume you know and can use the theorem that says dimension is well defined, meaning every basis has the same number of elements. Then in (i) you want to explicitly use the fact that a linearly independent set can be extended to a basis and when you extend you add vectors so the original number of vectors is less or equal than the dimension. If $m = n$ and you actually added anything when you extended to a basis then you would have bases of different sizes which is a contradiction, thus if $n = m$ it's already a basis.

For (ii) you basically use the same kind of arguments, I'll let you put it together.

So in summary your ideas are correct but you need to give more detail and be more careful when you explain them in a proof. More careful even then what I've written above. You should actually write down that you extend to a basis $\{v_1, \ldots, v_m, w_{m+1}, \ldots, w_{n'}\}$ with $m \leq n'$ and then cite the theorem about well definedness of dimension to get $n' = n$.

• Thanks for the input. Could you possibly expand on that theorem you mentioned? Commented Mar 16, 2013 at 16:11
• What do you mean expand? The precise statement of the theorem (in the finite dimensional case) is that if $V$ is a vector space and $\{v_1, \ldots, v_n\}$ and $\{w_1, \ldots, w_m\}$ are both bases of $V$ then $m = n$. The mere fact that you're talking about the dimension of a vector space means that you have almost surely covered this theorem in class.
– Jim
Commented Mar 16, 2013 at 16:13
• Sorry, I meant its inclusion in this answer. Your edit seems to cover it though. Commented Mar 16, 2013 at 16:14
• I'm getting a little confused. Should I begin by proving the 'basis extension theorem,' showing that $m \leq n'$ and then show $n'=n$ or would it work the other way around? Commented Mar 16, 2013 at 18:10
• You start by extending to a basis. If you don't do that first then there is no $n'$ defined for you to say that $n = n'$.
– Jim
Commented Mar 17, 2013 at 1:08