sign-flipping inverse - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T22:08:07Z http://mathoverflow.net/feeds/question/97653 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97653/sign-flipping-inverse sign-flipping inverse Felix Goldberg 2012-05-22T12:36:24Z 2012-05-23T02:26:23Z <p>Consider this matrix:</p> <p>$Z=\begin{bmatrix}23.9 &amp; -7 &amp; -17 \\ -7 &amp; 23.9 &amp; -17 \\ -17 &amp; -17 &amp; 33.9 \end{bmatrix}$</p> <p>Its inverse is entrywise negative (you can check...) and quite small in absolute value.</p> <p>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.</p> <p>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 $\widetilde{Z}$? </p> <p>Can you give a conceptual explanation? Is there a reference that targets this very specific issue?</p> http://mathoverflow.net/questions/97653/sign-flipping-inverse/97686#97686 Answer by Robert Israel for sign-flipping inverse Robert Israel 2012-05-22T17:07:40Z 2012-05-23T02:26:23Z <p>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 $\lambda$. </p>