# Basis of vectors.

by SherlockOhms
Tags: basis, vectors
 P: 309 So, I know that for a set of vectors to be a basis the set of vectors must be linearly independent and also must be a spanning set of vectors. So, they can't be parallel. I still feel that I'm not fully understanding what a basis is. Could someone explain to me, maybe with an example, what is a basis? Thanks.
 C. Spirit Sci Advisor Thanks P: 5,638 A basis allows you to represent any vector in the vector space as a unique linear combination of the basis vectors in said basis. This is the main utility of a basis. For example you can represent any vector in ##\mathbb{R}^{3}## as a unique linear combination of the basis vectors ##e_{1} = (1,0,0)^T, e_{2} = (0,1,0)^{T}, e_{3} = (0,0,1)^{T}##.
 Emeritus Sci Advisor PF Gold P: 4,500 In R2: {(1,0),(0,1)} is a basis. Given a vector (a,b) I can write it as a(1,0)+b(0,1) uniquely. {(1,0),(1,1)} is a basis. Given a vector (a,b) I can write it as (a-b)*(1,0)+b*(1,1) and this is the only way to do it. {(1,0)} is not a basis because it does not span the set of vectors {(1,0),(2,4),(-1,1)} is not a basis because the vectors are linearly dependent. I can write (2,4) = 4*(-1,1)+6*(1,0) In R3 some examples of bases: {(1,0,0),(0,1,0),(0,0,1)} {(1,0,0),(1,1,0),(1,1,1)} {(-1,14,12),(2,0,11),(65.3,114,-9)} The last one might not be immediately obvious that it is a basis, but the first and second one you should be able to prove. Some examples that are not bases: {(1,0,0),(14,12,10)}. This doesn't have enough vectors to span R3 (once you know the size of one basis is 3, all bases must be size 3), so it can't be a basis. After a little thought you should be able to explain why the vector (0,1,0) is not in the span of these two vectors. {(1,0,0),(0,1,0),(1,1,0)} has the right number of vectors, but is still not a basis. (1,1,0) is in the span of (1,0,0) and (0,1,0) so the vectors are not linearly independent, and they also are not a spanning set - (0,0,1) is not in the span. {(1,0,0),(0,1,0),(1,1,1),(1245,-9034,1234)} Has too many vectors, so cannot be a basis. You should be able to express (1245,-9034,1234) as a linear combination of the other three vectors. This one is at least a spanning set though, if that counts for anything
 P: 309 Basis of vectors. Thanks for that, lads. So, any arbitrary vector in R^3 can only be expressed in one way only in terms the 3 other vectors which form the basis of R^3? There's never an exception to this?
C. Spirit