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Math Help - Linear Transformations and Polynomials

  1. #1
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    Linear Transformations and Polynomials

    Consider the transformation of ax^2 + bx + c in P2 to absolute value of a in P0. Show that it does not correspond to a linear transformation by showing that there is no matrix that maps (a,b,c) in R^3 to absolute value of a in R.

    Any help is appreciated
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  2. #2
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    Quote Originally Posted by Shapeshift View Post
    Consider the transformation of ax^2 + bx + c in P2 to absolute value of a in P0. Show that it does not correspond to a linear transformation by showing that there is no matrix that maps (a,b,c) in R^3 to absolute value of a in R.

    Any help is appreciated
    Actually, I would think that the best way to show this is not a linear transformation would be to show that L(\alpha v) is not always equal to \alpha L(v). And do that by taking \alpha= -1.

    However, since you are asked to do this by showing that there is no matrix representing this linear transformation, take v_1= 1, v_2= x, and v_3= x^2 as basis for P2 and 1 as basis for P0. Then ax^2+ bx+ c= av_3+ bv_2+ cv_1. A standard way to construct a matrix for a linear transformation, given a basis for domain and range spaces, is to take that transformation of each of the basis vectors of the domain, in turn, and write the result as a linear combination of the basis vecotrs in the range. The coefficients are the columns of the matrix. Here, L(x^2)= L(1x^2+ 0x+ 0)=  1, L(x)= L(0x^2+ 1x+ 0)= 0 and L(1)= L(0x^2+ 0x+ 1)= 0. That is, if this were a linear transformation, its matrix in the standard bases would be the row matrix [1 0 0].

    But \begin{bmatrix}1 & 0 & 0\end{bmatrix}\begin{bmatrix}-1 \\ 0 \\ 0\end{bmatrix}= -1 while [tex]L(-x^2)= |-1|= 1.
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