Given a symmetric matrix $A$ and a vector $v$, will adding $v v^T$ to $A$ always increase or keep the eigenvalues of $A$ the same? The sum of the eigenvectors of $A + v v^T$ should be greater or equal to the sum of the eigenvectors of $A$ due to $tr(A + v v^T) = tr(A) + tr(v v^T)$, but one could increase one of the eigenvalues of the result matrix while decreasing another and still satisfy the sum increasing.
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4$\begingroup$ Are you assuming that $A$ is symmetric? In that case it should be true because $vv^T$ is positive semidefinite, but in general the eigenvalues need not even be real numbers. $\endgroup$– Noam D. ElkiesCommented Oct 14, 2016 at 2:19
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$\begingroup$ Yes, thank you for the clarification, I have edited the question. A is assumed symmetric. $\endgroup$– Christopher JohnsonCommented Oct 14, 2016 at 2:43
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1$\begingroup$ In that case, it was already answered (for the version where you subtract $vv^T$ and reduce the eigenvalues) in an earlier MO question, with the additional information that each eigenvalue increases to at most the next-largest one, attributed to Cauchy: mathoverflow.net/questions/193527/… $\endgroup$– Noam D. ElkiesCommented Oct 14, 2016 at 3:40
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2$\begingroup$ Each eigenvalue increases; this is immediate from the min-max principle: en.wikipedia.org/wiki/Min-max_theorem. Questions of this type are not really suited for this site. Please ask at MSE instead. $\endgroup$– Christian RemlingCommented Oct 14, 2016 at 3:59
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