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  1. #1
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    Base

    The set {1+t,t-t^2,1+2t-t^2} is a base of P2?
    Why?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    The set {1+t,t-t^2,1+2t-t^2} is a base of P2?
    Why?
    I think you mean a basis...

    First off, in order for this set to be a basis of P_2, it must have span P_2

    This means that a_1 (1+t)+a_2(t-t^2)+a_3(1+2t-t^2)=b_1t^2+b_2t+b_3 \implies (-a_2-a_3)t^2+(a_1+a_2+3a_3)t+a_1+a_3=b_1t^2+b_2t+b_3

    We see that the corresponding augmented matrix is \left[\begin{array}{ccc|c}0&-1&-1&b_1\\1&1&1&b_2\\1&0&1&b_3\end{array}\right]

    Getting it into RREF (I leave this for you to do), we see that \left[\begin{array}{ccc|c}1&0&0&b_1+b_2\\0&1&0&b_2-b_3\\0&0&1&-b_1-b_2+b_3\end{array}\right]

    Since the system has solutions (primarily a_1=b_1+b_2,~a_2=b_2-b_3, and a_3=-b_1-b_2+b_3), we see that \left\{1+t,t-t^2,1+2t-t^2\right\} spans P_2.

    Now, we need to show that these elements are linearly independent.

    a_1 (1+t)+a_2(t-t^2)+a_3(1+2t-t^2)=0

    This can be rewritten as (-a_2-a_3)t^2+(a_1+a_2+3a_3)t+a_1+a_3=0

    Thus, an augmented matrix can be constructed:

    \left[\begin{array}{ccc|c}0&-1&-1&0\\1&1&1&0\\1&0&1&0\end{array}\right]

    To test for linear independence, the easiest way is to evaluate \det\left(\left[\begin{array}{ccc}0&-1&-1\\1&1&1\\1&0&1\end{array}\right]\right)

    This equals \left|\begin{matrix}-1&-1\\1&1\end{matrix}\right|+\left|\begin{matrix}0&-1\\1&1\end{matrix}\right|=0+1=1\neq0

    Thus, they are linearly independent. Since this set spans P_2 and are linearly independent, they form a basis for P_2

    Does this make sense?

    --Chris
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  3. #3
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    thank you
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