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Vector Spaces & Subspaces

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://i26.lulzimg.com/274748.jpg [Broken]

    2. Relevant equations

    ??

    3. The attempt at a solution

    i dont even know how to start. lol.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 11, 2011 #2

    HallsofIvy

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    Are you sure you are in the right class then? If you honestly have "no idea where to start", if you don't know what a "subspace" is, what conditions a subspace must satisfy, or what the various symbols mean, you need more help than we can give. Talk to your teacher about this.
     
  4. Sep 11, 2011 #3

    Fredrik

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    Start by telling us the definition of a subspace. Then tell us what the notation in (a) means. Once you have done that, you shouldn't need us to tell you where to begin. The next step (the first step really) is obviously to check if W satisfies the conditions in the definition of a subspace of V.
     
  5. Sep 11, 2011 #4
    ok so i refreshed my memory a little, by looking at my notes. ive tried part (a), and got the following:
    condition (0) => A = [0 0;0 0] and A(Transpose) = [0 0;0 0], so satisfied
    condition(1) => A = [ a1 b1; c1 d1] and B = [ a2 b2; c2 d2]. A(Tran) + B(tran) must be equal to (A+B)(tran). since they are square, A(tran)= A, therefore satisfying condition (1)

    condition (2) => (c)*A must be equal to (c)*A(tran), and since A is square, they are equal, so condition 2 is satisfied.

    since all conditions are satisfied that means that W is a subspace of the Vectore Space V


    is that right?
     
  6. Sep 11, 2011 #5

    Fredrik

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    What you call condition(1) actually says that if A and B are in W, then so is A+B. So you need to show that for all A,B in W, [itex](A+B)^T=A+B[/itex]. It's not true that [itex]A^T=A[/itex] for all square matrices A. You should be able to find a counterexample of that very easily.
     
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