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Math Help - Matrix of a Linear Transformation

  1. #1
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    Question Matrix of a Linear Transformation

    L: P2 --> M22 given by

    L(ax^3 + bx^2 + cx + d) = Matrix
    -3a - 2c -b + 4d
    4b - c + 3d -6a - b + 2d


    For this linear transformation, what is the matrix for L with respect to the standard bases?

    Is it the 4x4
    -3 0 -2 0
    0 -1 0 4
    0 4 -1 3
    -6 -1 0 2
    ?

    Thanks,
    Last edited by Cantuba; April 28th 2012 at 04:47 PM.
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  2. #2
    MHF Contributor

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    Re: Matrix of a Linear Transformation

    Assume you mean P3 since that is your example. The standard basis for P3 is \{x^3, x^2, x, 1\} and, applying L to each of those,
    L(x^3)= \begin{bmatrix}-3(1)- 2(0) & -0+ 4(0)\\ 4(0)- 0+ 3(0) & -6(1)- (0)+ 2(0)\end{bmatrix}= \begin{bmatrix}-3 & 0  \\ 0 - 6\end{bmatrix}
    L(x^2)= \begin{bmatrix}-3(0)- 2(0) & -(1)+ 4(0) \\ 4(1)- 0+ 3(0) & -6(0)- 1+ 2(0)\end{bmatrix}= \begin{bmatrix}0 & -1 \\ 4 & -1\end{bmatrix}
    L(x)= \begin{bmatrix}-3(0)- 2(1) & -(0)+ 4(0) \\ 4(0)- 1+ 3(0) & -6(0)- 0+ 2(0)\end{bmatrix}= \begin{bmatrix}-2 & 0 \\ -1 &U 0\end{bmatrix}
    L(1)= \begin{bmatrix}-3(0)- 2(0)& -0+ 4(1) \\ 4(0)- 0+ 3(1) & -6(0)- 0+ 2(1)\end{bmatrix}= \begin{bmatrix}0 & 4 \\ 3 & 2\end{bmatrix}

    A basis for M_2 is
    [tex]\{M1= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix},M2= \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, M3=\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, M4=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
    s L(x^3)= -3M1_+ 0M2+ 0M3- 6M4
    so the first column is \begin{bmatrix}-3 \\ 0 \\ 0 \\ -6\end{bmatrix}.

    Yes, I get the same thing you do.
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