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:(adsbygoogle = window.adsbygoogle || []).push({});

[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.

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# Kernel of a quadratic form

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