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.



Spectral order of copositive matrices
The assertion is false. Here is how to construct a counterexample.


1. Let $A = XX^T + Y + Y^T$ where $Y \ge 0$ (elementwise)
2. 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 $\rho(B) = 9.8867$. 


