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Math Help - exams..help/.. subspace..dimension..

  1. #1
    n22
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    exams..help/.. subspace..dimension..

    Hi I need help with this .
    the set X={(x,y,z) which is an element of R^3 such that x=y=2z} is a subspace of R^3
    what is the dimension of X?
    Thanks.
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    Re: exams..help/.. subspace..dimension..

    Quote Originally Posted by n22 View Post
    Hi I need help with this .
    the set X={(x,y,z) which is an element of R^3 such that x=y=2z} is a subspace of R^3
    what is the dimension of X?
    Is this \{(1,1,0.5)\} the basis for X~?

    If it is then what is dimension of X~?
    Thanks from topsquark
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    Re: exams..help/.. subspace..dimension..

    x= y= 2z.

    So, whatever z is, y= 2z and x= 2z. Such a vector is (2z, 2z, a)= z(2, 2, 1). Now, what do you think a basis for that space is? What is its dimension?
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    n22
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    Re: exams..help/.. subspace..dimension..

    Quote Originally Posted by HallsofIvy View Post
    x= y= 2z.

    So, whatever z is, y= 2z and x= 2z. Such a vector is (2z, 2z, a)= z(2, 2, 1). Now, what do you think a basis for that space is? What is its dimension?
    Would the basis just now just be (2,2,1)<--since linearly independent .. and the dimension =1
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    Re: exams..help/.. subspace..dimension..

    Yes. And I had an obvious typo "Such a vector is (2z, 2z, a)= z(2, 2, 1)" should have been "Such a vector is (2z, 2z, z)= z(2, 2, 1)" but you understood that.

    Simlarly, if you have a subspace (x, y, z) satisfying only "y= z", x could be anything so such a vector would be (x, y, y)= (x, 0, 0)+ (0, y, y)= x(1, 0, 0)+ y(0, 1, 1). {(1, 0, 0), (0, 1, 1)} would be a basis and the dimension would be 2.
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