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MathGroup Archive 2005

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Re: Minimal maximum eigenvalue in closed form?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58371] Re: [mg58315] Minimal maximum eigenvalue in closed form?
  • From: Pratik Desai <pdesai1 at umbc.edu>
  • Date: Tue, 28 Jun 2005 21:57:03 -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
>
>  
>
Whoops! Forgot the most important part at the end
\!\(<< LinearAlgebra`MatrixManipulation`\[IndentingNewLine]
  \(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 - 2  x[1], 1 - x[2], 1 - x[3]}, {1,
            1 - x[5], \(-x[2]\) - x[6] + 1, 1 - x[2], 1 - 2  x[2],
            1 - x[3]}, {1, 1, 1 - x[6], 1 - x[3], 1 - x[3],
            1 - 2  x[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]};\)\[IndentingNewLine]
  mat1 = TakeMatrix[mat, {1, 1}, {6, 6}]\[IndentingNewLine]
  \(expr2 = \[Lambda]\ *IdentityMatrix[6] - mat1 // 
Det;\)\[IndentingNewLine]
  NSolve[expr2 == 0, {\[Lambda], x\_1, x\_2, x\_3, x\_4, x\_5, x\_6}]\)


Best regards
Pratik

-- 
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134



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