Given a real symmetric matrix $A$ and a vector $v$ of the same dimension we know that the eigenvalues of $A + vv^T$ are left interlaced by the eigenvalues of $A$.

  • But do we have any quantitative estimates of the amount of interlacing produced?

    Like as a function of $A$ and $v$ if we can say how much will the $k^{th}$ eigenvalue of $A+vv^T$ be ahead of the $k^{th}$ eigenvalue of $A$? (at least under some restrictions about the nature of $A$ and $v$?)

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    $\begingroup$ This seems to be the same question that you asked earlier here: mathoverflow.net/questions/193527/… $\endgroup$ – Christian Remling May 30 '15 at 1:03
  • $\begingroup$ @ChristianRemling They are not the same though its about the same concept. Here I am asking if we know of any bounds on the shifts that would happen. $\endgroup$ – user6818 May 30 '15 at 4:15
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    $\begingroup$ If you look at the trace you will see the total change of the eigenvalues. How this change distributes between the eigenvalues depends on where $v$ is with respect to the eigenvectors. $\endgroup$ – Brendan McKay May 30 '15 at 9:35
  • $\begingroup$ @BrendanMcKay True. But is that a way to quantify the shift in the eigenvalues in terms of A and v? $\endgroup$ – user6818 May 30 '15 at 17:52

It's easier to perform calculations in another direction: Instead of starting with $A$ and $v$, start with the spectrum of $A$ and the intended spectrum of $A+vv^\top$. One may use these spectra to calculate how much of $v$'s energy must lie in each eigenspace of $A$ (see for example Theorem 4 in this paper). Since your question concerns the "inverse" formula, I suspect the bounds you seek will quickly follow.

  • $\begingroup$ Is your linked paper making rank-1 updates? Can you may be help translate between the two terminologies? $\endgroup$ – Anirbit Jun 1 '15 at 3:05
  • $\begingroup$ @Anirbit: $F_{n+1}F_{n+1}^*=F_nF_n+f_{n+1}f_{n+1}^*$ $\endgroup$ – Dustin G. Mixon Jun 1 '15 at 12:37
  • $\begingroup$ So $A = F_nF_n$ and $v = f_{n+1}$ ? $\endgroup$ – Anirbit Jun 1 '15 at 15:18
  • $\begingroup$ @Anirbit: Indeed. $\endgroup$ – Dustin G. Mixon Jun 1 '15 at 15:26
  • $\begingroup$ Thanks for the clarification. You might have noticed that I had asked a related question here : mathoverflow.net/questions/208019/… $\endgroup$ – Anirbit Jun 1 '15 at 15:28

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