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Math Help - Dimension and basis

  1. #1
    Super Member Showcase_22's Avatar
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    Dimension and basis

    Find the dimension and a basis of the following vector spaces V over the given field K.
    i). V is the set of all vectors (\alpha,\beta,\gamma) in \mathbb{R}^3 with \alpha+4\beta-3\gamma=0; \ K= \mathbb{R}
    dim_{\mathbb{R}} \mathbb{R}^3=3.

    Is the basis just V=A(2,0,0)+B(0,1,0)+C(0,0,2)?
    I can't really see how to include the condition in a way better than this.

    ii). V is the set of all vectors (\alpha,\beta, \gamma) in \mathbb{R}^3 with \alpha+\beta=\gamma and \alpha-2\beta=-\gamma; \ k= \mathbb{R}.
    Once again dim_{\mathbb{R}} \mathbb{R}^3=3.

    However, i'm a little puzzled regarding this basis. Would it be:

    V=\alpha(1,0,0) + \beta(0,2,0) + \gamma (0,0,3)?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    dim_{\mathbb{R}} \mathbb{R}^3=3.

    Is the basis just V=A(2,0,0)+B(0,1,0)+C(0,0,2)?
    I can't really see how to include the condition in a way better than this.
    But you haven't included the condition at all. A general vector of that form would be (2A, B, 2C) and those do not satisfy (2A)+4(B)-3(2C)= 0. In order that (A, B, C) satisfy A+ 4B- 3C= 0 then you must have A= 3C- 4B. In particular, if C= 1 and B= 0, A= 3. <3, 0, 1> satisfies (3)+ 4(0)-3(1)= 0. Also if C= 0 and B= 1, A= -4. <-4, 1, 0> satisfies (-4)+ 4(1)- 3(0)= 0.
    The dimension is NOT 3, it is 2. And I know that because I just wrote a basis.

    Once again dim_{\mathbb{R}} \mathbb{R}^3=3.

    However, i'm a little puzzled regarding this basis. Would it be:

    V=\alpha(1,0,0) + \beta(0,2,0) + \gamma (0,0,3)?
    Once again, such a vector does NOT satisfy the condition you gave.
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