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Suppose $A$ is a real symmetric positive definite matrix with eigenvalues $\lambda_1,...,\lambda_n > 0$ (which we do not know). If one wants to know how many eigenvalues $A$ has above some limit $s \neq \lambda_i$, one can study the quadratic form defined by the matrix $(A-sI_n)$, diagonalise it to find the number of positive eigenvalues and use Sylvester's law of inertia. For example, if $(A-sI_n)$ has one positive eigenvalue, then $A$ has exactly one eigenvalue larger than $s$.

That is, the technique allows you to find the number of eigenvalues above some limit without actually computing the eigenvalues, and in order to work, the limit itself must not be an eigenvalue.

The literal translation of the name of this technique in my native language would be "spectral cleaving" or "spectral splitting". However, this doesn't seem to be the correct terminology in English. Any suggestions?

Ben Grossmann
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Étienne Bézout
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  • Where did you find this technique? Perhaps if we have a textbook name or author, someone will be better able to figure this out. – Ben Grossmann Aug 05 '16 at 19:33
  • @Omnomnomnom I found it in a textbook in my native language. The textbook is unlikely to be known internationally. – Étienne Bézout Aug 05 '16 at 19:43
  • Unlikely? Perhaps. Still, if you wouldn't mind posting the name, I think it would significantly increase the odds of your getting an answer here. It is unlikely that you're the only one on this site to speak your language. – Ben Grossmann Aug 05 '16 at 19:44
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    @Omnomnomnom I found an online document briefly describing the technique: http://dixon.hh.se/getc/LinSys/KvadratiskaFormer.pdf – Étienne Bézout Aug 05 '16 at 20:02
  • Which page specifically mentions the technique? – Ben Grossmann Aug 05 '16 at 20:11
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    @Omnomnomnom Sorry for the late reply. It's on page 7. – Étienne Bézout Aug 05 '16 at 22:43
  • I'm wondering if the term you are looking for is "spectral decomposition". – user115350 Aug 06 '16 at 17:44
  • @user115350 No, I don't think it is. The whole point of the technique is that you can say something about the location of the eigenvalues without actually computing them, using only Gaussian elimination to diagonalise a quadratic form. – Étienne Bézout Aug 06 '16 at 18:44

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I do not know your exact technique, but I know a rapid technique. You have a symmetric matrix, and are going to subtract off a multiple of the identity and ask how many of these eigenvalues are positive.

The point is that it is not necessary to know anything about the eigenvalues to solve $$ P^T (A-sI)P = D, $$ where we will perform about $n^2/2$ steps, resulting in $\det P = 1,$ then just count the positive entries in the diagonal matrix $P.$ I asked about an algorithm for this at reference for linear algebra books that teach reverse Hermite method for symmetric matrices

If, for example, all entries of $A-sI$ are integers, then all entries of $P$ will be rational numbers. Furthermore, it is usual (but not guaranteed) that $P$ is upper triangular with all diagonal entries $1.$

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Will Jagy
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    Thanks! In the method I'm referring to, you don't need to know the eigenvalues either, and the computation amounts to reducing $A-sI_n$ to row-echelon form so that you can find the diagonal elements. I think the calculations in our methods are virtually identical, but I'm only interested in the sign of the diagonal elements of $P^TAP$ for the sake of putting bounds on the eigenvalues. – Étienne Bézout Aug 05 '16 at 22:58
  • @ Will Jagy , about $O(n^2)$ operations, are you sure ? Each Step 2 Case 1 needs $O(k^2)$ operations. Step 3 uses $n-1$ times Step 2 Case 1. Then the number of operations is $\sum_{k\leq n}k^2=O(n^3)$. Am I wrong ? –  Aug 06 '16 at 12:46
  • @loupblanc It is my fault. I meant $(n^2-n)/2$ strict upper triangle positions to be set to zero. Each corresponds to one step $E^T A E,$ where $E$ is an elementary matrix. I suppose each "step" should be counted as a certain number of operations, possibly $n.$ I have never had any involvement with strict counting of operations, – Will Jagy Aug 06 '16 at 17:19