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Math Help - Change of basis?

  1. #1
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    Change of basis?

    Let B = {(1, -4), (-2, 9) and notice that B is a basis for R2. Therefore each (x,y) in R2 can be written as a linear combination of the vectors in B. If (x,y) = a(1,-4) + b(-2,9) then the coordinates of (x,y) relative to the basis B are said to be (a,b). (i.e. (x,y) = (a,b)B . Thus, associated with each (x,y) is another point in R2, (a,b). The function (sometimes called a transformation) that sends (x,y) to (a,b) is called a coordinate transformation. This transformation can be represented by a matrix. In other words, there is a matrix A such that the new coordinates (a,b) can be found by multiplying (x,y) on the left by A. Find A.



    I have an answer but I'm not sure if I'm doing this right.
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  2. #2
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    You have that

    \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}.

    Now you need to find A such that

    \begin{bmatrix} a \\ b \end{bmatrix} = A \begin{bmatrix}x \\ y \end{bmatrix}.

    But

    A \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = A \begin{bmatrix} x \\ y \end{bmatrix}.

    Therefore, A is the inverse of

    \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix}.
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  3. #3
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    I got
    \begin{bmatrix} 9x+2y \\ 4x+y \end{bmatrix} = x\begin{bmatrix} 9 \\ 4 \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \end{bmatrix}<br />
    as my answer.
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  4. #4
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    How did you arrive at that answer? I calculate

    A = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix},

    from which you get

    \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x + 4y \\ 2x + y \end{bmatrix}.
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    How did you arrive at that answer? I calculate

    A = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix},

    from which you get

    \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x + 4y \\ 2x + y \end{bmatrix}.
    Note, [LaTeX ERROR: Convert failed] indicates M inverse since I couldn't get M^(-1) to appear correctly.

    MA=B \ \ \ \  A=M^-B



    Where M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} =  \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix}



    B = \begin{bmatrix} x \\ y \end{bmatrix}

    M^- = \frac{1}{ad-bc}\*\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

    M^- = 1\*\begin{bmatrix} 9 & 2 \\ 4 & 1  \end{bmatrix}

    A = \begin{bmatrix} 9 & 2 \\ 4 & 1  \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x+2y \\ 4x+y \end{bmatrix} = x\begin{bmatrix} 9 \\ 4  \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \end{bmatrix}<br />
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  6. #6
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    So...what's right?
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  7. #7
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    Quote Originally Posted by chickeneaterguy View Post
    Note, [LaTeX ERROR: Convert failed] indicates M inverse since I couldn't get M^(-1) to appear correctly.

    MA=B \ \ \ \  A=M^-B



    Where M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} =  \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix}



    B = \begin{bmatrix} x \\ y \end{bmatrix}

    M^- = \frac{1}{ad-bc}\*\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

    M^- = 1\*\begin{bmatrix} 9 & 2 \\ 4 & 1  \end{bmatrix}

    A = \begin{bmatrix} 9 & 2 \\ 4 & 1  \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x+2y \\ 4x+y \end{bmatrix} = x\begin{bmatrix} 9 \\ 4  \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \end{bmatrix}<br />
    The question asked for a matrix. A, here, is vector, not a matrix.
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