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Change of basis matrix(linear algebra)

  1. Dec 19, 2011 #1
    Hi I'm stuck on this problem and I could not find similar examples anywhere.. any help would be greatly appreciated, thank you.

    1. The problem statement, all variables and given/known data
    Compute the change of basis matrix that takes the basis
    [itex]V1 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}[/itex] [itex]V2 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}[/itex]
    of R2 to the basis
    [itex]W1 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}[/itex] [itex]W2 = \begin{bmatrix} 3 \\ 7 \end{bmatrix}[/itex]
    I have done this first part, the change of basis matrix is [itex]A = \begin{bmatrix} 2 & 1 \\ -1 & 0 \end{bmatrix}[/itex]

    next part I don't quite know how to start:
    Consider v = V1 + 2(V2) [itex]\in[/itex] R2: Determine the column vector [itex]\begin{bmatrix} a \\ b \end{bmatrix}[/itex] which represents v with respect to the basis {W1, W2}

    3. The attempt at a solution

    Do I turn v1 into [itex]V1 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}[/itex] and V2 into [itex]V2 = \begin{bmatrix} 4 \\ 10 \end{bmatrix}[/itex] and then try and find a linear combination that gives me {W1, W2}?
     
  2. jcsd
  3. Dec 20, 2011 #2

    radou

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    Set up the linear combination and make it equal to your vector in order to find the coefficients a, b.
     
  4. Dec 20, 2011 #3

    vela

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    The fact that [itex]\vec{v} = 1\vec{v}_1 + 2\vec{v}_2[/itex] means that with respect to the [itex]\{\vec{v}_1,\vec{v}_2\}[/itex] basis,
    [tex]\vec{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}_{\{\vec{v}_1,\vec{v}_2\}}[/tex]Use the matrix A to convert the coordinates from one basis to the other.

    By the way, I don't think your matrix A is correct.
     
    Last edited: Dec 20, 2011
  5. Dec 20, 2011 #4
    Thank you for your answers. I made a mistake in the op. [itex]V1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}[/itex] and not (-1, 3) so my matrix A should be correct.
    Is multiplying A by (coefficients of v1, v2) [itex]v = \begin{bmatrix} 1 \\ 2 \end{bmatrix} [/itex] all I really need to do to?
    Then my answer is \begin{bmatrix} 4 \\ -1 \end{bmatrix}
     
  6. Dec 20, 2011 #5

    vela

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    It's easy enough to check. Is [itex]4\vec{w}_1-\vec{w}_2[/itex] equal to [itex]\vec{v}_1+2\vec{v}_2[/itex]?

    If your matrix is correct, then yes, that's all you have to do. That's why it's called a change of basis matrix. :wink:
     
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