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Coord vectors for polynomials

  1. Mar 6, 2005 #1
    Hi everyone,
    in this problem we are asked to get a coord vectors of polynomials with B as standard basis for P3 and then express one of the coord vectors as lin. combination of the others.
    So the set is this:
    {1-4x+4x^2+4x^3, 2-x+2x^2+x^3, -17 -2x-8x^2 + 2x^3}

    The way I was thinking is since basis is {1, x, x^2. x^3} the coord vectors of each polynomial is their coefficients. But then the answer in the textbook is that these are lin. dependent so that p3 = -p1 + 2p2, where p(i) is the polynomial in the set above. And I do not see how -17 = -1 +4 ????
    Any help is appreciated.
  2. jcsd
  3. Mar 6, 2005 #2


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    Well, first, I can tell you with confidence that -17 is NOT equal to -1+ 4 and therefore, p3 is NOT -p1+ 2p2, assuming that p1, p2, p3 are the polynomials above.

    Yes, If you take {1, x, x2,x3} as a basis for P3, then you can represent these as <1, -4, 4, 4> , <2,-1, 2, 1>, and
    <-17, -2, -8, 2>. To see IF they are dependent, you can write those as a matrix and row reduce. That's good practice but since these are given as polynomials, I think I would be inclined to treat this as simple algebra.

    The three polynomials are independent if and only the only way we can have Ap1+ Bp2+ Cp3= 0 (for all x) is if the number A, B, C are all 0.
    Ap1+ Bp2+ Cp3= A(1-4x+4x^2+4x^3)+ B(2-x+2x^2+x^3)+ C(-17 -2x-8x^2 + 2x^3)= (A+ 2B-17C)+ (-4A-B-2C)x+(4A+2B-8C)x2+(4A+B+2C)x3= 0. BECAUSE 1, x, x2, and x3 are independent, we must have A+ 2B- 17C= 0, -4A-B-2C= 0, 4A+2B-8C= 0, and 4A+ B+ 2C= 0, four equations for 3 unknown coefficients. Adding the second and third equations, B-10C= 0. Multiplying the first equation by 4 and adding to the second gives 7B- 138C= 0. Now multiply B- 10C= 0 by 7 and add to that last equation: 57C= 0 so C must equal 0. Then B= 0 and A= 0.

    Are you sure you have copied the problem correctly- those three polynomials ARE independent.
  4. Mar 6, 2005 #3
    Thanks for reply.
    I double-checked the example in the book, too bad I cannot find errata for it anywhere on the web.
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