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I'm reading one paper and on page 36 (48 in the pdf) it says:

Let d(s, i) be the (positive) diagonal terms that need to be subtracted from the matrix to make it negative semi-definite...

Could someone explain me why it's possible and how can I get the values of these terms?

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    $\begingroup$ I think you need to tell us more about the matrix and the context. Is the matrix Hermitian and positive definite to start? $\endgroup$
    – Geoff Robinson
    Jul 6, 2011 at 15:46
  • $\begingroup$ The paper is master thesis on Markov Random Fields and matrix elements are potentials of the variables. here is the thesis: stat.berkeley.edu/~pradeepr/paperz/thesis.pdf $\endgroup$
    – sbos
    Jul 6, 2011 at 15:53
  • $\begingroup$ My quote was from page 36 (48 in the pdf). $\endgroup$
    – sbos
    Jul 6, 2011 at 15:54
  • $\begingroup$ This might just be my lack of expertise in this area talking, but this seems like a question much better suited for direct email with the author. $\endgroup$
    – Cam McLeman
    Jul 6, 2011 at 16:00

1 Answer 1

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If $M$ is a Hermitian matrix, its eigenvalues $\lambda_j$ are all real. If you subtract $d$ from all the diagonal elements, you are changing $M$ to $M - d I$; if $d \ge \max_j \lambda_j$, all eigenvalues of $M - d I$ will be nonpositive so $M - d I$ will be negative semidefinite.

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  • $\begingroup$ That's nice, but what would be in more general case? $\endgroup$
    – sbos
    Jul 6, 2011 at 19:02
  • $\begingroup$ Ok, let M be symmetrical matrix $\endgroup$
    – sbos
    Jul 6, 2011 at 19:17
  • $\begingroup$ More general in what way? Perhaps subtracting $d_i$ from $M_{i,i}$ for each $i$? Then if $\min_i d_i \ge \max_j \lambda_j$, the new matrix will be negative semidefinite. $\endgroup$
    – Robert Israel
    Jul 7, 2011 at 18:11

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