MHB Spanning set of vectors question

[gucci1]

I am currently trying to understand linear independence and spanning sets. So the question that I have right now is, does a set of five linearly independent vectors always span F^5? Thanks for any help you can offer!

Edit: Sorry for the confusion, I really had no idea how to begin this problem. I guess I should have said that. I was pretty sure the answer was yes, but the reason why it is yes was a mystery.

 

[Opalg]

I am currently trying to understand linear independence and spanning sets. So the question that I have right now is, does a set of five linearly independent vectors always span F^5? Thanks for any help you can offer!

The answer is yes, and the reason is that the number of vectors in your set is the same as the dimension of the space.

The dimension of a vector space is the number of vectors in a basis, and the definition of a basis is that it is a set of vectors that satisfies these two properties:
(1) it is linearly independent,
(2) it spans the space.
To show that a set is a basis, you normally have to check that it satisfies both properties (1) and (2). But if you happen to know that the number of vectors in the set is the same as the dimension of the space, then you only have to check one of those two properties, and the other one will then automatically hold.

The reason for that is that there is a standard procedure for enlarging a linearly independent set so as to form a basis, by adding a suitable number of additional vectors. But there is a theorem (the basis theorem) which says that any two bases must contain the same number of elements. So if your linearly independent set started with the same number of elements as the dimension of the space, then the number of additional elements needed is zero. In other words, the original set was already a basis. Similarly, if you start with a set that spans the space, then you can always obtain a basis from it by discarding some of its elements. But if the spanning set has the same number of elements as the dimension of the space, then the number of elements to be discarded is zero, so the original set was already a basis.

 

[gucci1]

Thanks so much for the explanation, that really helped. I wasn't aware that it was so firmly established that the number of vectors in a basis is the same as the dimension of the space. I guess my teacher glossed over that fact :-/