(dim U+dim W)+dim V-dim (U+V+W) ........(*)
=dim (U+W)+dim (U\cap W)+dim V-dim (U+V+W)
=dim (U\cap W)+dim ((U+W)\cap V)
>=dim (U\cap W)
Since U,V,W are independent, and (*) is symmetric, so you get three inequalities.
Hi,
Have been trying to understand linear Algebra for a month now. Am unable to solve this:
If U,V,W are finite-dimensional subspaces of a real vector space, show that
dim U + dim V + dim W - dim(U+V+W) >= max{dim(U intersection V), dim(V intersection W), dim(U intersection W)}
Please help
He means the following(though I dont know the validity of the results):
Actually I dont know what U+V means. Is it the direct sum?
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And the validity of his idea depends on one concept,
But wiki says "The dimension of V ⊕ U is equal to the sum of the dimensions of V and U"...
So I am not sure
1. The sum U+V means the vector space consisting of elements of the form v+w where v in V and w in W, it is different from the direct sum . In the later case, we require if U,V are subspaces of a vector space. For instance, considering as a subspace of . But on the other hand, .
2. The "formula" was incorrect. The correct version is , . In general,
3. In my first post, I meant the sum of two vector subspaces U+V, not the direct sum.
Hope these help.