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Apr 13, 2017 at 12:58 history edited CommunityBot
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May 15, 2015 at 19:12 comment added Christian Remling @squattyroo: To address your actual comment, if $A=D^2+\alpha xx^t$ with $\alpha\in\mathbb R$, then the calculation from the other answer now gives $G_{\alpha}=F/(1+\alpha F)$, and thus the eigenvalues of $A$ are now the solutions of $F(\lambda)=-1/\alpha$ (above we're in the special case $\alpha=-1$). The (interlacing) bounds stay the same though (if $\alpha>0$, there's no eigenvalue below the spectrum of $D^2$ and instead there's one above the spectrum).
May 15, 2015 at 19:09 comment added Christian Remling @squattyroo: This (a rank-one perturbation with a coupling constant) is actually the situation I'm most familiar with. Somewhat unexpectedly, it's a pretty interesting subject; Barry Simon has a very readable review article on this, if you want to know more.
May 15, 2015 at 17:31 vote accept squattyroo
May 15, 2015 at 17:27 comment added squattyroo Very nice! Follow-up question: suppose we're looking at $D^2+\alpha xx^T$ for some $\alpha>0$. Then your analysis yields the largest eigenvalue is lower-bounded by $\alpha x_n^2$. It seems that it might be possible to analyze $F(\lambda)$ further and deduce a more interesting upper-bound than $\sum x_k^2$?
May 15, 2015 at 1:35 comment added Christian Remling I should also point out that the inequality $x_{n-1}^2\le\lambda\le x_n^2$ is just the interlacing property, I don't need to look at resolvents for this.
May 15, 2015 at 0:25 history edited Christian Remling CC BY-SA 3.0
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May 15, 2015 at 0:18 history edited Christian Remling CC BY-SA 3.0
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May 15, 2015 at 0:07 history edited Christian Remling CC BY-SA 3.0
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May 14, 2015 at 23:59 history answered Christian Remling CC BY-SA 3.0