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Kernel of a quadratic form

  1. Nov 20, 2009 #1
    Here is an interesting problem I came up with during my research. I first present a slightly simplified version. Let us the define component-wise the following bilinear symmetric form, returning a vector:

    [tex] a_i(u,v) = \frac{1}{2} (u^T A_i v - d_i) \;\;\; i=1 \ldots m [/tex]

    where [tex] u,v \in V = R^n, d_i = 1[/tex], and the [tex]A_i[/tex] are symmetric n-by-n matrices having constant vectors u=const in the kernel, i.e. rows and columns of [tex]A_i[/tex] add up to zero. Given A, I want to solve the following:

    Find [tex]u : a_i(u,u) = 0 \;\;\; \forall i=1 \ldots m,[/tex]

    Or, equivalently, I want to show that the only u satisfying the equation are constant vectors.

    I believe that a non-trivial u always exists in this simplified problem for a very large class of given A. The idea is to use the fact that A is diagonalizable, then it is possible to build u as a "sampling" proportional to (cos(t),sin(t)) on the eigenvector basis in order to pick up, for each component of a(), only two eigenvalues of the matrix and use the fact that cos^2+sin^2=1 to cancel the entries of d. However, I don't know if it's possible to implement a numeric test to check if this is true for a given A.
    Last edited: Nov 20, 2009
  2. jcsd
  3. Nov 20, 2009 #2
    To complete the statement, the more general problem consists in substituting d_i with a 2-by-2 identity matrix. It can be formulated as follows:

    a_{ij}(u,v) = \frac{1}{2} (u^T A_{ij} v - d_{ij}) \;\;\; i=1 \ldots m, \;\;\; j=1 \dots 4,

    where [tex]d_{i1} = d_{i2} = 1, d_{i3} = d_{i4} = 0[/tex]. Then:

    Find [tex]
    u : a_{ij}(u,u) = 0 \;\;\; \forall i=1 \ldots, m \;\;\; \forall j=1 \dots 4.

    My conjecture is that in this case a non-trivial solution does not exist unless a very special A is given. What test can I implement in Matlab or Mathematica? Newton's method is an overkill for just proving non-existence.
    Last edited: Nov 20, 2009
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