Do 4 Linearly Independent Vectors in R^4 Always Span the Space?

[lkh1986]

Homework Statement

You are given 4 vectors in R^4 which are linearly independent. Do they always span R^4?

Homework Equations

The Attempt at a Solution

Intuitively, I think the answer is yes. I know if I want to show they span R^4, I need to use the general terms, but all I can think of is the specific example case, i.e. standard basis for R^4, i.e. (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1). You see, the for vectors are linearly independence AND they span R^4 as well.

Unless someone wants to give me a hint to a counter-example? Thanks. :)

P.S. I also find this theorem: Is S is a set in R^n with n vectors, then S is a basis for R^n if either S spans R^n or S is linearly independent.

So, given 4 linearly independent vectors in R^4, by theorem, they form a basis, which implies they span R^4.

 

[HallsofIvy]

Yes. A "basis" for a vector space has three properties:
1) They span the space.
2) They are independent.
3) The number of vectors in the basis is equal to the dimension of the space.

And- any two of these is sufficient to prove the third. In your case, the vectors are independent and there are 4 of them so they span the space.

To prove that, assume there exist some vector, v, that is not in the span of the set of n independent vectors, where n is the dimension of the space. Then adding v to the set gives a set of n+1 vectors which are still independent. But one of the parts of the definition of "dimension" is that there cannot be any larger set of independent vectors.

 

[lkh1986]

Yes. A "basis" for a vector space has three properties:
1) They span the space.
2) They are independent.
3) The number of vectors in the basis is equal to the dimension of the space.

And- any two of these is sufficient to prove the third. In your case, the vectors are independent and there are 4 of them so they span the space.

To prove that, assume there exist some vector, v, that is not in the span of the set of n independent vectors, where n is the dimension of the space. Then adding v to the set gives a set of n+1 vectors which are still independent. But one of the parts of the definition of "dimension" is that there cannot be any larger set of independent vectors.

Thanks for the reply. :)