Question

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.

Answer 1

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 $\rho(B) = 9.8867$.

Answer 2

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