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I have solved an eigenvalue problem for matrix A which is orthogonal. I am trying to prove that the eigenvectors for matrix B is the same and find its eigenvalues. Matrix B is related to A as follows...

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All I know is that the eigenvalues of A^-1 are the inverse of eigenvalues of A. please help!

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  • $\begingroup$ So is $x$ any column of $B$? $\endgroup$ – Sean Roberson May 30 '17 at 20:58
  • $\begingroup$ sean i have updated $\endgroup$ – gamma1 May 30 '17 at 21:04
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Based on my interpretation, here's how it goes.

Let $(\lambda, y)$ be an eigenpair of $A$ (that is, $Ay = \lambda y$). Then

\begin{align*} By &= Ay + A^{-1} y + xIy \\ &= \lambda y + \frac{1}{\lambda} y + xy \\ &= \left( \lambda + \frac{1}{\lambda} + x \right) y \end{align*}

and so $\left( \lambda + \frac{1}{\lambda} + x, y \right)$ is an eigenpair for $B$.

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  • $\begingroup$ ah Ok, so y is your eigenvector. you initially state that it is an eigenvector of A, let B apply to it, and see what comes out? $\endgroup$ – gamma1 May 30 '17 at 21:29
  • $\begingroup$ Yeah, pretty much. $\endgroup$ – Sean Roberson May 30 '17 at 21:30

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