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Basis of vectors. 
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#1
May2313, 02:21 PM

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.



#2
May2313, 02:25 PM

C. Spirit
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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}##.



#3
May2313, 02:30 PM

Emeritus
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PF Gold
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In R^{2}:
{(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 (ab)*(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 R^{3} 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 R^{3} (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 


#4
May2313, 02:39 PM

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?



#5
May2313, 03:33 PM

C. Spirit
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#6
May2313, 04:12 PM

P: 309

Right, I've got it now. Thanks again.



#7
May2313, 11:45 PM

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P: 1,169

A basis also allows you to describe a transformation fully by describing its effect on basis vectors
(this is what a matrix does, w.respect to linear transformations). Given that in most cases (we're excluding here, e.g., vector spaces over finite fields) your spaces will contain infinitelymany vectors, this is a big plus. Re the uniqueness, assume (using a fixed basis for R^3 for definiteness; I think the generalization to other finitedimensional vector spaces is clear) the representation in terms of a fixed basis {v1,v2,v3} is not unique, so that we can write some v as: v=a1*v1+a2*v2+a3*v3 , and as: v=c1*v1+c2*v2+c3*v3 For triples (a1,a2,a3) , (c1,c2,c3) of scalars in your bases field; and neither triple has all zeros. Now, subtract one representation of v from the other, to get: (a1c1)*v1+(a2c2)*v2+(a3c3)*v3=0 Since the individual triples are not all zeros, this difference is a nonzero linear combination of basis vectors that gives you a zero. This is a contradiction of the assumption that {v1,v2,v3} is a basis. 


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