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

Correct reasoning about direct sums proof?

  1. Sep 18, 2007 #1
    Howdy everyone. I'm not very good at writing proofs, so I am wondering if someone can tell me if I'm even on the right page with this. I am not sure if I understand the idea correctly. The theorem goes as follows: Suppose V is a finite dim. vector space with subspaces U and W such that V is the direct sum of U and W. Let Z also be a subspace of V. Then Z cannot equal the direct sums of (i) the intersection of Z and U and (ii) the intersection of Z and W unless Z is a subspace of U or W.

    Here's my thoughts:

    Let "C" denote the direct sum of the intersections (i) and (ii) above. Then two conditions must hold:
    1) The intersection of the intersections (i) and (ii) must be the zero vector; and
    2) The sum of the intersections must equal V.

    The elements of C are given as {x|x is in Z and x is in U and x is in W}={0v}. Thus every x in Z that intersects U and W is in C. But we have that C={0v} since it is a direct sum. The only way that {0v}=(ZintersectU)intersect(ZintersectW) is if Z contains only the zero vector. Thus Z is the zero subspace, and it follows that Z must be a subspace of U and W.

    Yeah, let me know if that's right... I'm not sure that I am correct in asserting that Z contains only the zero vector. :rolleyes:
  2. jcsd
  3. Sep 18, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

  4. Sep 19, 2007 #3
    well it sounded reasonable to me... if C is the set of {(Z intersects U) intersects (Z intersects W)}. So for (Z intersects U) we have the set containing {x|x is in Z and x is in U} and similarly for (Z intersects W) we have the set {x|x is in Z and x is in W}.
    The direct sum would have all of these x too... wouldn't it? Although looking above it seems that I mixed up an intersection and direct sum. But even so, the intersection of the two intersections has to be the zero vector.
  5. Sep 19, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    Yes, which means your second sentence is irrelevant! So what if the intersection of those two things is {0}? C (the direct sum) will certainly contain this, but it definitely does not have to equal it.
  6. Sep 21, 2007 #5
    Maybe you've already proved it. If not, here's a suggestion.

    Put the statement into a form that'll shed some more light on the different options you have for proving it.
    The first thing I'd do is get rid of the "unless"; then you'll have a positive statement. It'll be an implication.
    Move its antecedent into your list of hypotheses (giving you more stuff to work with).
    Notice the antecedent's a disjunction (which suggests one proof method).
    Of course, there are other options too.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook