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Homework Help: Simple Linear algebra question but I have to have a brain freeze smh

  1. Feb 10, 2012 #1
    So I'm reading an example from my text on how to construct an isomorphism:

    Let V be in R^3 be defined by the single linear equation: x1 - x2 + x3 = 0 (those are suppose to be subscripts after the x's) and let W be in R^3 be the subspace defined by the single linear equation 2x1 + x2 - x3 = 0. Since dim(V) = dim(W) = 2, V and W are isomorphic. To construc an isomorphism we choose bases for V and W. A basis for v1 = (1,1,0) and v2 = (0,1,1). A basis for w1 = (1, -1, 1) and w2 = (-1/2, 1, 0)

    Question: How do I find the dim of V and W? Should be straight forward but I don't see it...smh, and how do they come up with those bases for V and W in the isomorphism?

  2. jcsd
  3. Feb 10, 2012 #2


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    hi trap101! :smile:
    every linear equation reduces the degrees of freedom by 1

    start with 3, apply one equation, result: 2 :wink:

    (ie, in n dimensions, one equation gives you an n-1 space, two simultaneous equations give you an n-2 space, etc)
    just choose numbers as simple as possible that satisfy the equation (and are independent) …

    use 1s and 0s if possible, if not then try 2s … :wink:
  4. Feb 10, 2012 #3
    Shouldn't I be able to form a linear combination from the standard basis vectors and solve for the dim in a matrix as well?......That was the route I was taking, but it might be useful to remember that fact from now on.
  5. Feb 10, 2012 #4


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    not following you :confused:
  6. Feb 10, 2012 #5
    this equation: x1 - x2 + x3 = 0 from the first part of my question, My intention was to write this out as a linear combination using the standard basis vectors in R^3, and then I wanted to find the basis from the reduced-echelon form of the matrix.

    Another question: How do those chosen bases relate to linear equations?
  7. Feb 10, 2012 #6


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    I'm surprised tiny-tim didn't say "(try using the X2 button just above the Reply box :wink:)"

    In terms of analytic geometry, the equations, x1 - x2 + x3 = 0 and 2x1 + x2 - x3 = 0 are equations of planes, so you can think of them as each having two dimensions.

    The vectors v1 = (1,1,0) and v2 = (0,1,1) each lie in the plane, V. Similarly, w1 = (1, -1, 1) and w2 = (-1/2, 1, 0) each lie in plane W.
    Last edited: Feb 10, 2012
  8. Feb 10, 2012 #7
    Another way you can look at it is this.
    The subspace you are talking about is this

    \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} with [itex] 2x_1 + x_2 - x_3 = 0[/itex]
    plugging in [itex] x_3 = 2x_1 + x_2[/itex]

    [itex]\begin{pmatrix} x_1 \\ x_2 \\ 2x_1 + x_2 \end{pmatrix} = x_1\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + x_2\begin{pmatrix} 0 \\1 \\1 \end{pmatrix}[/itex]

    Those are a base for the subspace and so the dimension is 2.
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