# Given a subspace S<=V, prove that there exists T<=V such that V=S⊕T.

1. Oct 23, 2012

### ashina14

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

V is a vector space

3. The attempt at a solution

If S is smaller than V then there exists a T such that S + T = V. OTHERWISE S = V. I'm not sure what assumptions am I making which I could break down to prove...

2. Oct 23, 2012

### hedipaldi

choose a basis A for T,complete it to a basis for V by adding a set B of vectors.Now show that the span of B is appropriate for T.

3. Oct 23, 2012

### ashina14

How do I show that the span of B is appropriate for T?

4. Oct 23, 2012

### ashina14

I've come up with this. Does this seem right?

if S is the empty space, the solution is obvious
if dim S >= 1 there is a base of vectors (ui) of S.
And there is a theorem who says that
the family of vectors (ui) can be completed
with a family of vectors (vj) so that
the union (ui) with (vj) is a basis of V
finally the subspace T generated by (vj) = (v1, v2, ...)
in the complementary space of S so that
V=S⊕T

5. Oct 23, 2012

### ashina14

i'm not sure if i need to quote the theorem

6. Oct 23, 2012

### hedipaldi

Assume that v is a non zero vector in the intersection of S and T and prove that this contradicts the linear independence of the vectors in the union of A and B.

7. Oct 23, 2012

### ashina14

How do I prove that? I don't see an obvious connection here

8. Oct 23, 2012

### Staff: Mentor

hedipaldi,
It is against Physics Forums rules to post complete solutions. You have received numerous warnings about this, and each comes with a private message to you. Apparently you aren't reading your PMs so I am posting something to you in this thread.

9. Oct 23, 2012

### ashina14

I didn't ask for a complete solution, I'm genuinely stuck. I'm not that acquainted with the unusual rules here as I don't come here often. Each of your warning is about a separate issue and I don't repeat the same mistake again. I would appreciate if you understand my position.

10. Oct 23, 2012

### hedipaldi

I know this rule and indeed i answered by hints.however it was not understood so i tried to help more.I understand that this is unwanted and i will obey the rules of the forum.
sorry'
Hedi

11. Oct 23, 2012

### ashina14

Thanks for the help anyway :)

12. Oct 23, 2012

### Staff: Mentor

My post was addressed to hedipaldi, not you, ashina14. See above.