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Math Help - prove a vector subspace

  1. #1
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    prove a vector subspace

    Can someone please help me with this question, ?

    I know i first need to prove that it is empty,

    how do i show that the zero vector is in U?

    i know I have to take two elements of U and add them and see if I still am in the same space, but really not sure how to?

    any help appreciated.
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  2. #2
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    Re: prove a vector subspace

    To show the 0-matrix is in U, you need to verify that:

    A0B = B0A, can you do this?

    For your next question, suppose that:

    AXB = BXA and AYB = BYA.

    You must show that A(X+Y)B = B(X+Y)A.

    Start by expanding the LHS to get:

    A(X+Y)B = (AX + AY)B = AXB + AYB. Do the same for the RHS. What can you say about: A(X+Y)B - B(X+Y)A ?
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  3. #3
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    Re: prove a vector subspace

    Quote Originally Posted by Deveno View Post
    To show the 0-matrix is in U, you need to verify that:

    A0B = B0A, can you do this?

    For your next question, suppose that:

    AXB = BXA and AYB = BYA.

    You must show that A(X+Y)B = B(X+Y)A.

    Start by expanding the LHS to get:

    A(X+Y)B = (AX + AY)B = AXB + AYB. Do the same for the RHS. What can you say about: A(X+Y)B - B(X+Y)A ?
    so

    AXB + AYB = BXA + BYA

    AB( X+Y) = BA(X+Y) ?


    does this prove it?
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  4. #4
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    Re: prove a vector subspace

    You don't prove something by starting out with what you're trying to prove. You start with what you're GIVEN.

    We want to prove that if X and Y are in U, then X+Y is in U. Maybe this isn't true, we need to find out. What is our criterion for deciding if X,Y are in U? That AXB = BXA, and AYB = BYA. So that is where we start.

    So the question is: is it THEN true that:

    A(X+Y)B = B(X+Y)A?

    Now A(X+Y)B = (A(X+Y))(B) (associative law for matrix multiplication)

    = (AX+AY)B (distributive law for matrix muultiplication)

    = AXB + AYB (distributive law again).

    Since AXB = BXA (given),

    = BXA + AYB ....can you continue?

    It is NOT true that:

    AXB + AYB = BXA + BYA implies AB(X+Y) = BA(X+Y), in general, you can't "treat matrices like numbers" and "change the order of multiplication" (bring stuff "in front"). AB and BA are usually two different matrices.
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  5. #5
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    Re: prove a vector subspace

    And if X in U, show kX in U, k a scalar.
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  6. #6
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    Re: prove a vector subspace

    Quote Originally Posted by Hartlw View Post
    And if X in U, show kX in U, k a scalar.
    Towards this end, it might be helpful to recall that:

    kX = (kI)X and that the matrices kI DO commute with any matrix, so k(AXB) = (kI)(AXB) = A(kI)(XB) = A((kI)(X))B = A(kX)B.
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