Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook