Spectral order of copositive matrices It is curious to know whether the following assertion is ture or not?
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.
For positive definite matrices class and nonnegative (entrywise) matrices class, this is obviously true.
 A: The assertion is false. Here is how to construct a counterexample.


*

*Let $A = XX^T + Y + Y^T$ where $Y \ge 0$ (elementwise)

*Let $B = XX^T$


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.
However, if you try the above recipe to construct $A$ and $B$, then you get the following counterexample (via Matlab again) very rapidly.
$ X = \begin{pmatrix}
  -1.8393& -0.9342\\\\
  1.7632 & 1.6479
\end{pmatrix}$
$Y = \begin{pmatrix}
  1.9949& 2.0663 \\\\
  2.3393& 0.1889
\end{pmatrix}
$
$A = \begin{pmatrix}
  8.2456 & -0.3770\\\\
  -0.3770& 6.2024
\end{pmatrix}
$
$B =
\begin{pmatrix}
  4.2558 &-4.7826\\\\
 -4.7826 &5.8247
\end{pmatrix}$
Here, we have $\rho(A) = 8.3130$ and $\rho(B) = 9.8867$. 
A: 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.
Practically Yes: If $A$,$B^{T}$ (or, $A^{T},B$) have the Perron-Frobenius property and $A \leq B$, then $\rho(A) \leq \rho(B)$.
But: If $A \leq C \leq B$ and $A$,$B^{T}$ have the Perron-Frobenius property, then $\rho(C)$ can fall below $\rho(A)$.
Both the theorem and an example for the second statement can be found in: Dimitrios Noutsos, On Perron–Frobenius property of matrices having some negative entries, Linear Algebra and its Applications 412 (2006) 132–153
