Spectral order of copositive matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:49:55Z http://mathoverflow.net/feeds/question/68891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68891/spectral-order-of-copositive-matrices Spectral order of copositive matrices Sunni 2011-06-27T00:22:19Z 2012-05-03T12:16:40Z <p>It is curious to know whether the following assertion is ture or not?</p> <p>If $A-B$ and $B$ are copositive matrices (implying $A$ is copositive) of the same size, then $\rho(A)\ge \rho(B)$, where $\rho$ means the spectral radius.</p> <p>For positive definite matrices class and nonnegative (entrywise) matrices class, this is obviously true.</p> http://mathoverflow.net/questions/68891/spectral-order-of-copositive-matrices/68900#68900 Answer by S. Sra for Spectral order of copositive matrices S. Sra 2011-06-27T02:39:03Z 2011-06-27T02:39:03Z <p>The assertion is false. Here is how to construct a counterexample.</p> <ol> <li>Let $A = XX^T + Y + Y^T$ where $Y \ge 0$ (elementwise)</li> <li>Let $B = XX^T$</li> </ol> <p>Then, by construction $A$ is a copositive matrix (sum of semidefinite plus symmetric nonnegative matrix), and $B$ is copositive too (because it is semidefinite). Moreover, $A-B$ is also copositive because it is just a symmetric nonnegative symmetric.</p> <p>However, if you try the above recipe to construct $A$ and $B$, then you get the following counterexample (via Matlab again) very rapidly.</p> <p>$X = \begin{pmatrix} -1.8393&amp; -0.9342\\ 1.7632 &amp; 1.6479 \end{pmatrix}$</p> <p>$Y = \begin{pmatrix} 1.9949&amp; 2.0663 \\ 2.3393&amp; 0.1889 \end{pmatrix}$</p> <p>$A = \begin{pmatrix} 8.2456 &amp; -0.3770\\ -0.3770&amp; 6.2024 \end{pmatrix}$</p> <p>$B = \begin{pmatrix} 4.2558 &amp;-4.7826\\ -4.7826 &amp;5.8247 \end{pmatrix}$</p> <p>Here, we have $\rho(A) = 8.3130$ and $\rho(B) = 9.8867$. </p> http://mathoverflow.net/questions/68891/spectral-order-of-copositive-matrices/95855#95855 Answer by Felix Goldberg for Spectral order of copositive matrices Felix Goldberg 2012-05-03T12:16:40Z 2012-05-03T12:16:40Z <p>It is interesting to consider the same question for matrices with the Perron-Frobenius property (that may have negative entries). The answer is: practically yes, but.</p> <p>Practically Yes: If $A$,$B^{T}$ (or, $A^{T},B$) have the Perron-Frobenius property and $A \leq B$, then $\rho(A) \leq \rho(B)$.</p> <p>But: If $A \leq C \leq B$ and $A$,$B^{T}$ have the Perron-Frobenius property, then $\rho(C)$ can fall below $\rho(A)$.</p> <p>Both the theorem and an example for the second statement can be found in: Dimitrios Noutsos, <em>On Perron–Frobenius property of matrices having some negative entries</em>, Linear Algebra and its Applications 412 (2006) 132–153</p>