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Post Closed as "Not suitable for this site" by Christian Remling, Wolfgang, Stefan Waldmann, Stefan Kohl♦, Alexey Ustinov
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Christopher Johnson
Christopher Johnson
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Given a squaresymmetric 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.

Given a square 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.

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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Christopher Johnson
Christopher Johnson
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Is adding an outer product to a matrix guaranteed to not decrease the eigenvalues?

Given a square 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.