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Sum of Two Subspaces

  1. Jul 6, 2009 #1
    This is an example that I am a little confused by:

    [tex]U={(x,0,0)\in\mathbf{F}^3:x\in\mathbf{F}}\text{ and }W={(0,y,0)\in\mathbf{F}^3:y\in\mathbf{F}}[/tex]

    Then
    [tex]U+W={(x,y,0):x,y\in\mathbf{F}[/tex]

    Okay, I get that. Now it says that U is defined the same as above but now let
    [tex]W={(y,y,0)\in\mathbf{F}^3:y\in\mathbf{F}}[/tex]

    Then the sum of U and W is the same as given above. Why is that? What is happening to that y that is in the "x" position?


    Perhaps I am confusing the definition of the sum of two lists with the sum of two subspaces.
     
    Last edited: Jul 6, 2009
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  3. Jul 6, 2009 #2

    Dick

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    If (a,b,0) is any point there is an x and y that solve it in the first case and there is also an x and y that solve it in the second case. Isn't this so? x and y are ANYTHING, right? Who cares what happened to the y?
     
  4. Jul 6, 2009 #3
    Sorry. I guess I am getting lost in all of these definitions.

    I guess I don't really understand what it means to 'Add' two subspaces...

    See I have it in my head that I should be getting something like (x+y,y,0) ...but I know that is not right.

    It says in my book that the "sum of two subspaces is the set of all possible sums of the elements of the two subspaces."

    I guess I really don't understand what that means.:confused:
     
  5. Jul 7, 2009 #4

    Office_Shredder

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    So if you have W in the first case:

    U+W is the set of all vectors of the form u+w where u is in U and w in W. SO u+w = (x,0,0) + (0,y,0) = (x,y,0) where x and y are arbitrary.

    For the second case, u=(x',0,0) for any x', and w = (y',y',0) for any y'. Then u+w = (x'+y',y',0). Now, we want to show that (x,y,0) is in U+W for arbitrary x and y. Given y, set y'=y and now we need that x'+y' =x i.e. x' = x-y. Hence we see we can find x' and y' such that u+w = (x,y,0) as required.

    Basically, the point is that the set of all vectors of the form (x+y,y,0) is
    1) two dimensional
    2) contained in the set of all vectors of the form (x,y,0)

    So they must be the same subspace
     
  6. Jul 7, 2009 #5
    Office_Shredder: That is starting to sink in a little. Thanks for the explanation. I am finally at the end of this gruesome 1st chapter and can move onto the exercises. I think I will get a better understanding of these concepts by doing them.
     
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