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gorica

  • 3 years ago

Let W, S, T be subspaces of finite-dimensional vector space such that S∩T=S∩W S+T=S+W, and W is subset of T. Prove that W=T.

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  1. Edutopia
    • 3 years ago
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    ..

  2. abb0t
    • 3 years ago
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    Definition of Composition?

  3. malevolence19
    • 3 years ago
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    I'm not sure of a formal proof but if you think about it the only way that S+T=S+W is if they are the same. If they were not the same that would mean there is an element of T not in W or an element of W not in T. If not you'd have something like {1,2}+{3,4,5}={1,2}+{3,4,6} implies {1,2,3,4,5}={1,2,3,4,6} which is not true obviously. It is still true that their intersections can be equal. {1,2} intersect {3,4,5} is the same as {1,2} intersect {3,4,6} (in this case the empty set). I think that the fact of their intersections with S are equal is superfluous information but I didn't care for proofs too much.

  4. gorica
    • 3 years ago
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    actually, I have to prove that T is subset of W, using what is given, and that will imply that T=W since I already have given that W is subset of T.

  5. AddemF
    • 3 years ago
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    Consider any arbitrary element of T, call it t. We want to show that t is in W. Suppose that t is not in W and let's try to get a contradiction. If t is not in W then it is still in S+T which is the same as S+W and so there exist some s in S and w in W such that t = s+w. But since we know W is a subset of T, then this w is in T. For that reason, let's re-name it t1, to show that it is actually in T. That means t = s+t1. And since the intersection of S and T is a subspace, it is closed under addition. Therefore s+t1 is an element of the intersection of S and T, and since this is the same thing as t, then t is in the intersection of S and T. But the intersection of S and T is the same as the intersection of S and W. This means t is in W. This contradicts our initial assumption. Therefore t is in W. By definition of set inclusion, T is a subset of W.

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