# Linear Algebra: Spans and Dimensions

1. Feb 25, 2013

### Millacol88

1. The problem statement, all variables and given/known data
Given v1, v2 ... vk and v, let U = span {v1, v2 ... vk} and W = span {v1, v2 ... vk, v}. Show that either dim W = dim U or dim W = 1 + dim U.

3. The attempt at a solution
I'm not really sure where to start. If I knew that {v1, v2 ... vk} was linearly independent, then it would be easy to prove. But I'm not given anything about linear independence so I'm at a loss.

2. Feb 25, 2013

### LCKurtz

What if you reduce $\{v_1,v_2,...,v_k\}$ to a basis set for U to work with?

3. Feb 25, 2013

### Millacol88

Okay, so if I reduced the generating set for U to a basis, I would know its linearly independent, but then where would I go with it?

4. Feb 25, 2013

### Karnage1993

Do the same for W. Note, W is also spanned by v which may or may not be represented as a linear combination of the other vectors (this is why there are two possible cases for the dimensions).

5. Feb 25, 2013

### HallsofIvy

Consider two cases:
1) v is in U.
2) v is not in U.

6. Feb 25, 2013

### Millacol88

Ok, so assuming we reduce the set in U to a basis: then if v is in U the given generating set for W is linearly dependent. Then v would be removed from this set, making it linearly independent. This implies that U and V share a basis, and thus their dimensions are the same. If v is not in U, then the given generating set is linearly independent, and thus is a basis for W. Therefore dim W = dim U + 1. Does this make sense?

7. Feb 26, 2013

### HallsofIvy

I would think it was obvious that if v is in U then adding v to U doesn't change the span: span(U)= span(V) so their dimensions are the same. It does NOT follow that "If v is not in U, then the given generating set is linearly independent" but it is not necessary. What ever the dimension of U, if v is not in U, adding it increases the dimension by 1.