# Re: [R] Gaussian elimination - singular matrix

From: Spencer Graves <spencer.graves_at_pdf.com>
Date: Wed, 27 Jun 2007 19:21:32 -0700

RSiteSearch("generalized inverse", "fun") produced 194 hits for me just now, including references to the following:

Ginv {haplo.stats}
ginv {MASS}
invgen {far}
Ginv {haplo.score}

At least some of these should provide a solution to a singular system. You could further provide side constraints as Moshe suggested, but I suspect that 'ginv', for example, might return the minimum length solution automatically.

Hope this helps.
Spencer Graves

Moshe Olshansky wrote:
> All the nontrivial solutions to AX = 0 are the
> eigenvectors of A corresponding to eigenvalue 0 (try
> eigen function).
> The non-negative solution may or may not exist. For
> example, if A is a 2x2 matrix Aij = 1 for 1 <=i,j <=2
> then the only non-trivial solution to AX = 0 is X =
> a*(1,-1), where a is any nonzero scalar. So in this
> case there is no non-negative solution.
> Let X1, X2,...,Xk be all the k independent
> eigenvectors corresponding to eigenvalue 0, i.e. AXi =
> 0 for i = 1,2,...,k. Any linear combination of them,
> X = X1,...,Xk, i.e. a1*X1 + ... + ak*Xk is also a
> solution of AX = 0. Let B be a matrix whose COLUMNS
> are the vectors X1,...,Xk (B = (X1,...,Xk). Then
> finding a1,...,ak for which all elements of X are
> non-negative is equivalent to finding a vector a =
> (a1,...,ak) such that B*a >= 0. Of course a =
> (0,...,0) is a solution. The question whether there
> exists another one. Try to solve the following Linear
> Programming problem:
> max a1
> subject to B*a >= 0
> (you can start with a = (0,...,0) which is a feasible
> solution).
> If you get a non-zero solution fine. If not try to
> solve
> min a1
> subject to B*a >= 0
> if you still get 0 try this with max a2, then min a2,
> max a3, min a3, etc. If all the 2k problems have only
> 0 solution then there are no nontrivial nonnegative
> solutions. Otherwise you will find it.
> Instead of 2k LP (Linear Programming) problems you can
> look at one QP (Quadratic Programming) problem:
> max a1^2 + a2^2 + ... + ak^2
> subject to B*a >= 0
> If there is a nonzero solution a = (a1,...,ak) then X
> = a1&X1 +...+ak*Xk is what you are looking for.
> Otherwise there is no nontrivial nonnegative solution.
>
> --- Bruce Willy <croero_at_hotmail.com> wrote:
>
>
>> I am sorry, there is just a mistake : the solution
>> cannot be unique (because it is a vectorial space)
>> (but then I might normalize it)
>>
>> can R find one anyway ?
>>
>> This is equivalent to finding an eigenvector in
>> fact> From: croero_at_hotmail.com> To:
>> r-help_at_stat.math.ethz.ch> Date: Wed, 27 Jun 2007
>> 22:51:41 +0000> Subject: [R] Gaussian elimination -
>> singular matrix> > > Hello,> > I hope it is not a
>> too stupid question.> > I have a singular matrix A
>> (so not invertible).> > I want to find a nontrivial
>> nonnegative solution to AX=0 (kernel of A)> > It is
>> a special matrix A (so maybe this nonnegative
>> solution is unique)> > The authors of the article
>> suggest a Gaussian elimination method> > Do you know
>> if R can do that automatically ? I have seen that
>> "solve" has an option "LAPACK" but it does not seem
>> to work with me :(> > Thank you very much>
>>
>>
> _________________________________________________________________>
>
>> Le blog Messenger de Michel, candidat de la Nouvelle
>> Star : analyse, news, coulisses… A découvrir !> >
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>>
>>
> _________________________________________________________________
>
>> Le blog Messenger de Michel, candidat de la Nouvelle
>> Star : analyse, news, coulisses… A découvrir !
>>
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>>
>>
>>> ______________________________________________
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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Thu 28 Jun 2007 - 02:39:50 GMT

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