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Subspaces and dimension

  1. Mar 13, 2008 #1
    subspaces and dimension!!!!

    Consider two subspaces V and W of R^n ,where V is contained in W.
    Why is dim(V)<= dim(W).....?
    "<=" less than or equal to
  2. jcsd
  3. Mar 13, 2008 #2
    the only way you get dim(V) = dim(W) is if V=W, if V is strictly contained in W then, then there must be some vector in W that is not in V, let this be w. Now let (v_1,v_2,...v_r) be a basis of V (now I assumed dim(W)=r). Clearly v_1,v_2,...v_r are in W because it containes V, and because w is not in V it must be independent of v_1,v_2,...v_r, so
    (v_1,v_2,...v_r,w) is a linearly independent set in W, so dim(W) is at least r+1 (could be more).
  4. Mar 13, 2008 #3


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    Just because I can't resist "putting my oar in", I'll echo mrandersdk: If V is a subspace of W, then any vector any V is also a vector in W. Any basis for W spans V and so any basis for V cannot be larger than a basis for W. Since the dimension of a space is the size of a basis, ....

    By the way, do you see why saying "subspace V is a subset of subspace W" (they are both subspaces of some vector space U) is the same as saying "V is a subspace of W"?
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