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

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Homework Help Overview

The discussion revolves around proving the existence of a subspace T such that the vector space V can be expressed as the direct sum of a subspace S and T, denoted as V = S ⊕ T. The context involves concepts from linear algebra, specifically related to vector spaces and subspaces.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the dimensions of S and V, with some suggesting the completion of a basis for T to form a basis for V. Questions arise regarding the appropriateness of the span of certain vectors and the implications of linear independence.

Discussion Status

The discussion includes attempts to clarify the necessary conditions for the existence of T and the implications of various assumptions. Some participants express uncertainty about specific steps and theorems, while others emphasize adherence to forum rules regarding the provision of complete solutions.

Contextual Notes

There are indications of confusion regarding forum rules about providing hints versus complete solutions, which may impact the flow of the discussion. Participants also question the assumptions made in their reasoning and the need for formal proofs.

ashina14
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Homework Statement



V is a vector space


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...
 
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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.
 
How do I show that the span of B is appropriate for T?
 
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
 
i'm not sure if i need to quote the theorem
 
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.
 
How do I prove that? I don't see an obvious connection here
 
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.
 
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
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
Thanks for the help anyway :)
 
  • #12
Mark44 said:
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.

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

The rules are here: https://www.physicsforums.com/showthread.php?t=414380.
 
  • #13
hedipaldi said:
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
I'm glad to hear that!
 

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