Suppose I have a hermitian matrix $X$ such that the difference between consecutive non-zero eigenvalues of $X$ (when the eigenvalues are arranged in increasing order) is at most $\delta$. We consider a new matrix $X' = \Pi X \Pi$, where $\Pi$ is a projector (or a matrix with eigenvalues either $0$ or $1$). Then is it true that the difference between consecutive non-zero eigenvalues of $X'$ (when the eigenvalues are arranged in increasing order) is also at most $\delta$ (or even a constant times $\delta$) ?
Gap between eigenvalues of a matrix under projection
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linear-algebra
matrices
eigenvalues-eigenvectors
1 Answers
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No. Example: $X=diag(0,\delta,2\delta,3\delta)$, $\Pi=diag(1,1,0,1)$. Then $\Pi X\Pi=diag(0,\delta,0,3\delta)$.