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Re: Minimal maximum eigenvalue in closed form?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg58370] Re: [mg58315] Minimal maximum eigenvalue in closed form?
*From*: Pratik Desai <pdesai1 at umbc.edu>
*Date*: Tue, 28 Jun 2005 21:57:02 -0400 (EDT)
*References*: <200506280913.FAA05092@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Paul Abbott wrote:
>Here is an interesting exercise: compute the minimal maximum eigenvalue
>of the matrix (arising in a semidefinite programming problem)
>
> mat =
> {
> {1, 1 - x[4], 1 - x[4], 1 - x[4], 1, 1},
> {1 - x[4], 1, 1 - x[5], -x[1] - x[5] + 1, 1 - x[5], 1},
> {1 - x[4], 1 - x[5], 1, 1 - x[1] - x[6], 1 - x[2] - x[6], 1 - x[6]},
> {1 - x[4], 1-x[1] -x[5], 1-x[1] -x[6], 1 - 2x[1], 1 - x[2], 1 - x[3]},
> {1, 1 - x[5], -x[2] - x[6] + 1, 1 - x[2], 1 - 2x[2], 1 - x[3]},
> {1, 1, 1 - x[6], 1 - x[3], 1 - x[3], 1 - 2x[3]}
> };
>
>in closed form. This is reminiscent of the sort of problems given in the
>SIAM 100 digit challenge, see
>
> mathworld.wolfram.com/Hundred-DollarHundred-DigitChallengeProblems.html
>
>Numerically, the answer is 1.5623947722331...
>
>It can be shown that the exact answer can be expressed as the root of a
>6th order polynomial. Does anyone have an elegant way of obtaining the
>solution (and also the values of x[1] through x[6])?
>
>Cheers,
>Paul
>
>
>
Hi Paul,
Here is my somewhat lame attempt. Although I am not clear what minimal
maximum value means, perhaps Least Upper Bound and what is a closed form
solution in this case since you are calculating, in effect ,the roots of
a 6th order polynomial. Using NSolve I get the following,
Clear[mat,mat1]
<<LinearAlgebra`MatrixManipulation`
mat={{1,1-x[4],1-x[4],1-x[4],1,1},{1-x[4],1,1-x[5],-x[1]-x[5]+1,1-x[5],
1},{1-x[4],1-x[5],1,1-x[1]-x[6],1-x[2]-x[6],1-x[6]},{1-x[4],
1-x[1]-x[5],1-x[1]-x[6],1-2x[1],1-x[2],1-x[3]},{1,
1-x[5],-x[2]-x[6]+1,1-x[2],1-2x[2],1-x[3]},{1,1,1-x[6],1-x[3],
1-x[3],1-2x[3]}}/.{x[1]->Subscript[x,1],x[2]->Subscript[x,2],
x[3]->Subscript[x,3],x[4]->Subscript[x,4],x[5]->Subscript[x,5],
x[6]->Subscript[x,6]};
mat1=TakeMatrix[mat,{1,1},{6,6}]
expr2=\[Lambda] *IdentityMatrix[6]-mat1//Det;
Best regards
Pratik
--
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134
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