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Math Help - Finding a basis

  1. #1
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    Finding a basis

    If possible, find a basis n = {n1, n2, n3} of P2(R) such that

    [2 + 5x + 4x^2]n = [1,2,3] , [1 + x + x^2]n = [4,1,2] and

    [x + x^2]n = [3, -5, 1]

    Basically, from what I understand, we have Ax = b.

    I can find a basis, for one of these, but not all three.

    like, [a,b,c] * [2,5,4]^t = [1,2,3]^t
    [d,e,f]
    [g,h,i]

    0, 1, -1
    1, 0, 0
    -1, 1, 0 so, [0,1,-1], [1,0,1] [-1,0,0] would be a basis for the first one, but the others... I'm not sure...

    Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by CaughtARattleSnake View Post
    [2 + 5x + 4x^2]n = [1,2,3]

    What does the above equality mean?. Coordinates of 2 + 5x + 4x^2 on n ?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    If the answer to my question is yes then, denoting:

    B=\{1,x,x^2\},\;B'=\{2 + 5x + 4x^2,1 + x + x^2,x + x^2\}

    you have enough information to find P,Q\in\mathbb{R}^{3\times 3} such that for every f(x)\in P_2(\mathbb{R}) :

    [f(x)]_B=P[f(x)]_{B'},\quad[f(x)]_n=Q[f(x)]_{B'}

    This implies [f(x)]_B=PQ^{-1}[f(x)]_n and finding PQ^{-1} , you determine inmediately n .
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