# Linear algebra proof and vectors

1. Mar 6, 2009

### bcjochim07

1. The problem statement, all variables and given/known data

a) Prove that if u1, ... um are vectors in Rn , S = {u1,u2,...uk} and T = {u1,...uk, uk+1,...um} then span(S) $$\subseteq$$span(T).

b) deduce also that if Rn = span(S), then Rn=span(T)

2. Relevant equations

3. The attempt at a solution
I think I got part a:

the span S is represented by linear combination c1u1 + c2u2 + ... ckuk
and the span T is represented by the linear combination
c1u1 + c2u2 + ... ckuk +... cmum
and since span(S) is contained in span(T)

span(S)$$\subseteq$$span(T)

Does this look alright?

b) part b is giving me trouble

since span(S) = Rn, the entire set of linear combinations of the vectors {u1,u2,...uk} in set S forms a plane.

since span(S) is a subset of (T) which spans Rn,
span(T) = Rn.

I don't feel like this right at all, and I really can't visualize in my head what's going on here. Could somebody please help me? Thanks.

2. Mar 7, 2009

### lurflurf

for b)
We know span(T) contains R^n.
How do we know it can not contain anything else?

3. Mar 7, 2009

### maze

Algebraically you are on the right track. Just remember these definitions:
1) Span(u1, u2, ..., ui) = {a1u1 + a2u2 + ... + aiui | a1, a2, ..., ai in R}

2) A is a subset of B if every x in A is also in B.

3) Sets A and B are equal if A is a subset of B and B is a subset of A.

For some intuition, it may be helpful to consider what this looks like in R3. Here are some pictures:

http://img242.imageshack.us/img242/1087/spanliner3.png [Broken]

http://img243.imageshack.us/img243/8680/spanplaner3.png [Broken]

http://img243.imageshack.us/img243/9002/spanblockr3.png [Broken]

Last edited by a moderator: May 4, 2017