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Consider this matrix:

$Z=\begin{bmatrix}23.9 & -7 & -17 \\\\ -7 & 23.9 & -17 \\\\ -17 & -17 & 33.9 \end{bmatrix}$

Its inverse is entrywise negative (you can check...) and quite small in absolute value.

Now, the eigenvalues of $Z$ are $-0.1,30.9,50.9$ and if I take the matrix $\widetilde{Z}=Z+(0.1+\epsilon)I$ its inverse flips over into being entrywise positive and very large.

Now, I understand that the poor conditioning of $\widetilde{Z}$ is responsible for the large entries in absolute magnitude - but why the bizarre behaviour of the signs of the entries of $\widetilde{Z}$?

Can you give a conceptual explanation? Is there a reference that targets this very specific issue?

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  • $\begingroup$ In working with 0-1 matrices, one observes this sign inversion phenomenom with certain extremal matrices. (I never looked for how the signs on the inverse changed when perturbing the matrix.) Since I was looking for maximum determinant on a bounded set, I might suggest that it bears some resemblance to Lagrange multipliers for determining optima, and that choice of direction contributes (or not) to the effect. Gerhard "Ask Me About System Design" Paseman, 2012.05.22 $\endgroup$
    – Gerhard Paseman
    Commented May 22, 2012 at 15:07
  • $\begingroup$ Is it significant that the diagonal entries have the form $(a^2 - b^2 -1)/(a-b)$ and $(a^2 + b^2 + 1)/(a-b)$ where $a=17$ and $b=7$? $\endgroup$
    – Barry Cipra
    Commented May 22, 2012 at 16:07
  • $\begingroup$ @Barry Cipra, Probably not, the matrix originally came from the stuff I am working on which has a lot of structure, but presumably the phenomenon is a general one. But thanks for noticing. $\endgroup$
    – Felix Goldberg
    Commented May 23, 2012 at 10:21

1 Answer 1

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By Cramer's rule, if $A$ is an $n \times n$ matrix $(A-tI)^{-1} = \dfrac{Adj(A-tI)}{\det(A-tI)}$. So if $\lambda$ is a simple real eigenvalue of the real matrix $A$, but all the first minors of $A - \lambda I$ are nonzero, then the signs of all entries of $(A-tI)^{-1}$ will flip as $t$ crosses $\lambda$.

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  • $\begingroup$ Robert, I am sure I am missing something: the determinant flips sign, but what about the minors in the adjugate? How do you how their signs behave? Thanks a million. $\endgroup$
    – Felix Goldberg
    Commented May 23, 2012 at 10:23
  • $\begingroup$ They are also polynomials, so their signs may flip somewhere, just (under the assumption they're nonzero at $\lambda$) not at $\lambda$. In your example, the roots of entries of the adjugate are $16.9$, $30.9$, $5263/70$, and $28.9 \pm \sqrt{314}$, none of which happen to be between $0$ and $-0.1$. $\endgroup$
    – Robert Israel
    Commented May 23, 2012 at 17:05

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