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Math Help - Linearly independence and dependence of polynomials

  1. #1
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    Linearly independence and dependence of polynomials

    Hi

    How do we work out if the vectors in P2 are linearly dependent or independent.

     <br />
p1= 2-x + 4x^2; p2= 3 + 6x +2x^2; p3= 2+10x-4x^2<br />

    Thanks
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  2. #2
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    The most direct way is to use the definition of "linear indepdence" and "linear dependence":
    If there exist numbers, a_1, a_2, ..., a_n, not all 0, such that a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n= 0, where that "0" is the 0 vector, then the vectors v_1, v_2, ..., v_n are dependent, if not, then they are independent.

    So you start by setting up the linear combination a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n which, here, is a_1p_1+ a_2p_2+ a_3p_3= a_1(2- x+ 4x^2)+ a_2(3+ 6x+ 2x^2)+ a_3(2+ 10x- 4x^2)= 0 where that "0" is the 0 function, f(x)= 0 for all x.

    Now you can "combine like terms" and use the fact that if a polynomial is 0 for all x then all coefficients must be 0 to get three equations to solve for a, b, and c. Since all three equations will be "= 0", the have the obvious solution a= b= c= 0. If that is only solution, then the vectors are independent. It that solution is not unique, if there exist other, non-zero, solutions, then they are dependent.

    (You don't have to actually find a, b, and c. You can use the fact, if you aready know it, that a system of equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.)
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    The most direct way is to use the definition of "linear indepdence" and "linear dependence":
    If there exist numbers, a_1, a_2, ..., a_n, not all 0, such that a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n= 0, where that "0" is the 0 vector, then the vectors v_1, v_2, ..., v_n are dependent, if not, then they are independent.

    So you start by setting up the linear combination a_1v_1+ a_2v-2+ \cdot\cdot\cdot+ a_nv_n which, here, is a_1p_1+ a_2p_2+ a_3p_3= a_1(2- x+ 4x^2)+ a_2(3+ 6x+ 2x^2)+ a_3(2+ 10x- 4x^2)= 0 where that "0" is the 0 function, f(x)= 0 for all x.

    Now you can "combine like terms" and use the fact that if a polynomial is 0 for all x then all coefficients must be 0 to get three equations to solve for a, b, and c. Since all three equations will be "= 0", the have the obvious solution a= b= c= 0. If that is only solution, then the vectors are independent. It that solution is not unique, if there exist other, non-zero, solutions, then they are dependent.

    (You don't have to actually find a, b, and c. You can use the fact, if you aready know it, that a system of equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.)
    Thank you so the determinant of that is -32, therefore it doesnt have a unique solution therefore its linearly dependent.
    Last edited by adam_leeds; August 4th 2010 at 07:01 AM.
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  4. #4
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    Quote Originally Posted by adam_leeds View Post
    Thank you so the determinant of that is -32, therefore it doesnt have a unique solution therefore its linearly dependent.
    Is this right?

    I did the determinant of

    2 -1 4
    3 6 2
    2 10 -4
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