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Linear Algebra - Quadratic polynomial to Matrix

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Examining the answers of the previous two questions, write the quadratic polynomial f(x1,x2,x3)=x1x2−6x22+3x2x3−3x23 in the form
    f(x1,x2,x3)=[x1x2x3]A[x1x2x3]<-this last group is a column matrix
    where A is a symmetric matrix.


    2. Relevant equations
    Matrix multiplication


    3. The attempt at a solution
    So the previous problems had me start with the matrix, then multiply by the row then column vectors to get a polynomial. Then here it wants me to work backwards to get matrix A.

    I started with:
    x1x2−6x22+3x2x3−3x23.
    Grouped like terms:
    (x1x2) + (−6x22 + 3x2x3) + (−3x23)
    Took out an x1, x2, x3 form each grouping, respectively. That gives me the second step of the problem:
    [(x2) (−6x2 3x3) (−3x3)] [x1 x2 x3] <-again, column matrix
    So, working backwards, I now have to find the values of the elements of matrix A so when multiplied by row vector [x1 x2 x3] will result in the output of [(x2) (−6x2 3x3) (−3x3)].
    In other words, [x1][# # #](<-column) will equal [x2].
    So I end up getting matrix A =
    [ 0 0 0
    1 -6 0
    0 3 -3]

    I know A11 = 0 and A22 = -6 are correct. But when I input the rest of the matrix in they homework system tells me I'm wrong.
    Could anyone look through this and see if I made a mistake somewhere? Or maybe the system has the wrong key and my answer is correct? Any insight and help is greatly appreciated.

    Thanks
     
  2. jcsd
  3. Sep 11, 2011 #2

    lanedance

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    that matrix is not symmetric for a start
     
  4. Sep 11, 2011 #3

    lanedance

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    now say you have found a matrix B that is not symmetric, but gives you teh correct polynomial. you can always write it in terms of a symmetric (S) and skew-symmetric (P) part
    B = S+P

    consider the polynomial
    p(x) = xTBx = xTSx + xTPx =

    as p(x) is a scalar function, it will be equal to its transpose
    p(x) = p(x)T = (xTSx)T + (xTPx)T = (xTST x)] + (xTPTx)
    = (xTS x) - (xTPx)

    which implies xTPx = 0 that only the symmetric part of the matrix is important in defining the polynomial

    so if you are happy you matrix is correct (have you checked it gives the right result?)

    you can find the symmetric and anti symmetric parts as follows

    B = S+P
    S = (B+BT)/2
    P = (B-BT)/2
     
  5. Sep 12, 2011 #4
    Thank you for the explanation lanedance. I didn't know about symmetric matrices.
     
  6. Sep 12, 2011 #5

    lanedance

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    no worries - just to add, for most polynomials there is actually an infinite way they can be written interms of a matrix x^TBx, however if you restrict yourself to symmetric matrix , which appear to be the natural choice based on teh above, then is is one unique solution
     
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