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Let $A$ be a real symmetric invertible matrix and $b$ a real non-zero vector. Consider the problem of finding a real number non-zero $\lambda$ and a real valued vector $x$ such that $$Ax=\lambda x + b.$$

How can I numerically and efficiently solve this problem?

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    There are couple of trivial (silly?) solutions with $\lambda = 0, 1.$ If $\lambda = 0,$ then $Ax = b,$ or $x = A^{-1} b.$ If $\lambda = 1,$ then $Ax = x + b,$ or $x = (A - I)^{-1} b.$2012-05-26
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    I understand my question may sound silly. There is only a finite number of eigenvalues and I am looking at finding numerically a non-eigenvalue (λ=1 can be an eigenvalue, for exemple take A=I).2012-05-26
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    One thing to note is that you get a (unique) solution for *every* choice of $\lambda$ that isn't an eigenvalue. (and maybe even some solutions where $\lambda$ is an eigenvalue) If it's okay to diagonalize, that makes it easy to find many solutions.2012-05-26

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