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Related post Find the eigenvector associated with the smallest eigenvalue, not smallest in magnitude

I tried to find the smallest eigenvalue for a generalized eigenvalue problem A c= \lambda B c (the default option is by absolute value) by

a = {{1, 2, 0}, {2, 5, 3}, {0, 3, 9}}; b = {{1, 5, 3}, {5, 2, 0}, {3, 5, 0}};

n = 5; dimension = 3; Eigenvalues[N[{a, b}], 3]

-n + Eigenvalues[ N[{ a + n IdentityMatrix[dimension], b + n IdentityMatrix[dimension] }, 32 ] ]

It does not work. How to find smallest eigenvalue for a generalized eigenvalue problem?

P.S. In my case (before shifting N* Identity matrix), all eigenvalues are real.

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user26143
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  • One approach is to get an a priori estimate of largest in magnitude, and shift along positive real axis with a multiple of identity matrix so that the most negative eigenvalue becomes also the largest. This will only work if there are no complex ones that become larger still. – Daniel Lichtblau Jan 22 '15 at 16:13
  • I tried to shift by n IdentityMatrix[dimension] in both matrices $a$ and $b$, seems this prescription only works in eigenvalue not generalized eigenvalue problem.. – user26143 Jan 22 '15 at 17:35
  • If your eigenvalues are complex, how would you define the "smallest"--magnitude? – David G. Stork Jan 22 '15 at 18:07
  • I guess I had not realized the general case would behave differently. I sent a link to the question to someone more knowledgeable, so maybe a better answer will emerge. – Daniel Lichtblau Jan 22 '15 at 19:02
  • Sorry for confusing, in my case, all eigenvalues are real (before shifting N* Identity matrix). – user26143 Jan 23 '15 at 02:32
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    Does this help? – Szabolcs Jan 23 '15 at 04:51
  • @Szabolcs Yes, it works (with a=-a) for a and b are real and symmetric. b is positive, which is the actual case I am interested in. Thank you so much! – user26143 Jan 23 '15 at 09:52

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