Spectral order of copositive matrices - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T19:49:55Zhttp://mathoverflow.net/feeds/question/68891http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/68891/spectral-order-of-copositive-matricesSpectral order of copositive matricesSunni2011-06-27T00:22:19Z2012-05-03T12:16:40Z
<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#68900Answer by S. Sra for Spectral order of copositive matricesS. Sra2011-06-27T02:39:03Z2011-06-27T02:39:03Z<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& -0.9342\\
1.7632 & 1.6479
\end{pmatrix}$</p>
<p>$Y = \begin{pmatrix}
1.9949& 2.0663 \\
2.3393& 0.1889
\end{pmatrix}
$</p>
<p>$A = \begin{pmatrix}
8.2456 & -0.3770\\
-0.3770& 6.2024
\end{pmatrix}
$</p>
<p>$B =
\begin{pmatrix}
4.2558 &-4.7826\\
-4.7826 &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#95855Answer by Felix Goldberg for Spectral order of copositive matricesFelix Goldberg2012-05-03T12:16:40Z2012-05-03T12:16:40Z<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>