[SciPy-Dev] Generalized eigenproblem with rank deficient matrices
Charles R Harris
charlesr.harris@gmail....
Sun Sep 4 12:12:01 CDT 2011
On Sun, Sep 4, 2011 at 10:55 AM, Charles R Harris <charlesr.harris@gmail.com
> wrote:
>
>
> On Sun, Sep 4, 2011 at 10:09 AM, Nils Wagner <nwagner@iam.uni-stuttgart.de
> > wrote:
>
>> On Sun, 4 Sep 2011 09:29:19 -0600
>> Charles R Harris <charlesr.harris@gmail.com> wrote:
>> > On Sun, Sep 4, 2011 at 7:53 AM, Nils Wagner
>> ><nwagner@iam.uni-stuttgart.de>wrote:
>> >
>> >> Hi all,
>> >>
>> >> how can I solve the eigenproblem
>> >>
>> >> A x = \lambda B x
>> >>
>> >> where both matrices are rank deficient ?
>> >>
>> >
>> > I'd do eigh and transform the problem to something like:
>> >
>> > U * A * U^t * x= \lambda D * x
>> >
>> > where D is diagonal. Note that the solutions may not be
>> >unique and \lambda
>> > can be arbitrary, as you can see by studying
>> >
>> > A = B = array([[1, 0], [0, 0]])
>> >
>> > Where there are solutions for arbitrary \lambda.
>> >Likewise, there may be no
>> > solutions under the requirement that x is non-zero:
>> >
>> > A = array([[1, 1], [1, 0]]),
>> > B = array([[1, 0], [0, 0]])
>> >
>> > The usual case where B is positive definite corresponds
>> >to finding extrema
>> > on a compact surface x^t * B *x = 1, but the surface is
>> >no longer compact
>> > when B isn't positive definite. Note that these cases
>> >are all sensitive to
>> > roundoff error.
>> >
>> > Chuck
>>
>> Hi Chuck,
>>
>> I am only interested in the real and complex
>> eigensolutions.
>> The complex eigenvalues appear in pairs a_i \pm \sqrt{-1}
>> b_i sind A and B are real.
>> How can I reject infinite eigenvalues ?
>> Both matrices, A and B, are indefinite.
>>
>>
> It depends on the particular problem. In general, the solution to the
> generalized eigenvalue problem starts by making a variable substitution that
> reduces B to the identity matrix, usually by using a Cholesky factorization,
> i.e., B = U^t U, y = U x. This can still be done in the (numerically)
> indefinite case but Cholesky won't be reliable and that is why I suggested
> eigh. Note that the problem we are trying to solve is
> finding the extrema of x^t A x subject to the constraint x^t B x = 1,
> \lambda is then a Lagrange multiplier. I assume A and B are both symmetric?
> Anyway, in terms of y, the problem then reduces to finding extrema of y^t
> U^t^{-1} A U^{-1} y subject to y^t D y = 1, where D is diagonal and has ones
> along part of the diagonal, zeros for the remainder. In the usual case, D is
> the identity. The trick is then to divide y into two parts, one for where D
> is one (u), another for the rest (v), so that y = [u v]. If you are lucky,
> the v can be solved in terms of u using the transformed A, and things will
> reduce to an eigenvalue problem for u. If the original A was symmetric, so
> will be the reduced problem. If you can't solve v as a function of u, then
> you can reduce things further, but it is possible that at some point there
> is no solution.
>
> I don't have practical experience with this sort of problem with indefinite
> B, so I can't tell you much more than that. I assume you've googled for
> relevant documents.
>
> It occurs to me that even if A and B aren't symmetric, that you can still
bring B to the desired form using the svd.
Chuck
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