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Math Help - Change of co-ordinates

  1. #1
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    Change of co-ordinates

    I have the following
    B:=[p_0,p_1,p_2] denote the natural ordered basis for P_2(R)
    C:=[h_1,h_2,h_3] where h_1(x)=7x^2+2x+3; h_2(x)=4x^2+5x+7 and
    h_3(x)=-9x^2+x+1

    Construct the change of coordinate matrix A which converts C-coordinates to B-coordinates.

    Now my first thought was the rows in C are given by the columns in B, however when going through the calculations to construct f_c, in a later calculation. Do i have to find an expression for p_0, p_1 and p_2 using the equations?

    many thanks in advance.
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  2. #2
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    Re: Change of co-ordinates

    presumably, you mean B = \{1,x,x^2\}, that is, p_0(x) = 1, p_1(x) = x, p_2(x) = x^2. suppose we have a vector in P_2(\mathbb{R}) ginve in C-coordinates, that is:

    v = c_1h_1 + c_2h_2 + c_3h_3 or v = [c_1,c_2,c_3]_C. in other words,

    h_1 = [1,0,0]_C,h_2 = [0,1,0]_C,h_3 = [0,0,1]_C.

    now if we had such a matrix A. then:

    A([h_i]_C) would be its i-th column. but that would just be the i-th basis vector of C in B-coordinates. so, for example,

    A([h_1]_C) = [h_1]_B = [3,2,7]_B, since:

    h_1(x) = 3 + 2x + 7x^2 = 3p_0(x) + 2p_1(x) + 7p_2(x)

    convince yourself that a linear combination of the basis vectors in C is taken to the same linear combination of the images of the basis vectors in B-coordinates, because a matrix is a linear mapping.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Change of co-ordinates

    Another approach: by a well known theorem if B=\{u_1,\ldots,u_n\},\;B=\{u'_1,\ldots,u'_n\} are basis of a vector space V and \begin{Bmatrix} u'_1=a_{11}u_1+\ldots +a_{1n}u_n\\...\\ u'_n=a_{n1}u_1+\ldots +a_{nn}u_n\end{matrix} then, [x]_B=P\;[x]_{B'} where P=\begin{bmatrix} a_{11}  & \ldots & a_{n1}\\ \vdots&&\vdots \\ a_{1n} &\ldots & a_{nn}\end{bmatrix}

    In our case, we immediately get [x]_B=\begin{bmatrix}{3}&{7}&{\;\;1}\\{2}&{5}&{\;\;1}  \\{7}&{4}&{-9}\end{bmatrix}[x]_C
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  4. #4
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    Re: Change of co-ordinates

    Quote Originally Posted by Deveno View Post
    presumably, you mean B = \{1,x,x^2\}, that is, p_0(x) = 1, p_1(x) = x, p_2(x) = x^2. suppose we have a vector in P_2(\mathbb{R}) ginve in C-coordinates, that is:

    v = c_1h_1 + c_2h_2 + c_3h_3 or v = [c_1,c_2,c_3]_C. in other words,

    h_1 = [1,0,0]_C,h_2 = [0,1,0]_C,h_3 = [0,0,1]_C.

    now if we had such a matrix A. then:

    A([h_i]_C) would be its i-th column. but that would just be the i-th basis vector of C in B-coordinates. so, for example,

    A([h_1]_C) = [h_1]_B = [3,2,7]_B, since:

    h_1(x) = 3 + 2x + 7x^2 = 3p_0(x) + 2p_1(x) + 7p_2(x)

    convince yourself that a linear combination of the basis vectors in C is taken to the same linear combination of the images of the basis vectors in B-coordinates, because a matrix is a linear mapping.
    Yes A([h_1]_C) = [h_1]_B = [3,2,7]_B, A([h_2]_C) = [h_2]_B = [7,5,4]_B and A([h_3]_C) = [h_3]_B = [1,1,-9]_B
    so can we say that M=A (inverse), where
    M=3 7 1
    2 5 1
    7 4 -9
    and just find the inverse of M to find our matrix A?
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  5. #5
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    Re: Change of co-ordinates

    no, the matrix:

    \begin{bmatrix}3&7&1\\2&5&1\\7&4&-9\end{bmatrix}

    is the matrix A, it changes C-coordinates into B-coordinates.

    if you have a non-standard basis, and you want to change coordinates from the non-standard basis to the standard basis,

    you write down the matrix whose columns are the non-standard basis vectors coordinates in the standard basis,

    that is the "change of coordinate" matrix. if you want to change from B-coordinates to C-coordinates, THEN you would take the inverse of A.
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