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Dimension of vector spaces

  1. Nov 21, 2008 #1
    Hi all!

    I´m trying to prove following two inequalities but I somehow got stuck:

    U, W are subspaces of V with dimV = n

    1) dimV >= dim(U+W)

    2) dim(U+W)>=dimU and dim(U+W)>=dimW

    Could you give me some hints?

    thanks in advance!
  2. jcsd
  3. Nov 21, 2008 #2


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    Just notice that if U and W are subspaces of V, then U+W is too. So pick a basis, and see what you can do with it. The same principle works for the second part also
  4. Nov 21, 2008 #3
    ok, let u_1,.....,u_n be a basis of U and w_1,....,w_m be a basis of W.

    Every vector x of U+W can be therefore expressed as a lin combination of some u_1,..u_n,w_1,..,w_n , which implies that u_1,..u_n,w_1,..,w_n is a generating set of U+W. But we know that the basis of U+W must be a minimal generating set, so letting p be its dimension this yields: p<=m+n, or dim(U+W)<= dimU + dimV. Now we consider w element of W, which we also find in W+U, but cannot be spanned by u_1,...,u_n only, so dimW<=dim(W+U), where W is a subspace of W+U. but W+U is also a subspace of V, so in the end, we have

    dim(U+W)<= dimV

    is this a regular proof, or I did something wrong?

    and another question: dimU + dimW should not necesserily be smaller or equal to dimV, should it?
  5. Nov 21, 2008 #4
    You're making it much more complicated than it needs to be.

    Since U + W is a subspace of V, it follows immediately that dim(U + W) ≤ dim V. The same idea works for the second part.

    As for your last question, dim U + dim W is not necessarily less than or equal to dim V (take U = V and W = V, for example).

    Here's a bonus question for you: Show that dim U + dim W = dim(U + W) + dim(U ∩ W).
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