# Basis of vectors.

1. May 23, 2013

### SherlockOhms

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. May 23, 2013

### WannabeNewton

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. May 23, 2013

### Office_Shredder

Staff Emeritus
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

4. May 23, 2013

### SherlockOhms

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. May 23, 2013

### WannabeNewton

By definition, any vector in R^3 MUST have a unique linear expression in terms of a chosen basis for R^3. Now the basis that I wrote above isn't the only basis you can pick for R^3. There are many bases you can pick so a given vector in R^3 has many different expansions in terms of different bases.

6. May 23, 2013

### SherlockOhms

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

7. May 23, 2013

### Bacle2

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 infinitely-many 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 finite-dimensional 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:

(a1-c1)*v1+(a2-c2)*v2+(a3-c3)*v3=0

Since the individual triples are not all zeros, this difference is a non-zero linear
combination of basis vectors that gives you a zero. This is a contradiction of the
assumption that {v1,v2,v3} is a basis.