A Metzler matrix is a matrix $A$ such that $a_{ij} \geq 0$ for any indices $i \neq j$ (i.e. non-diagonal entries are non-negative). A stable Metzler matrix is a Metzler matrix that has eigenvalues with negative real part.
Let $A$ be a $n \times n$ stable Metzler matrix and let $B$ be a $n \times n$ Metzler matrix with $b_{ii}=0$. If $a_{ij}-b_{ij}>0$, is the matrix $A-B$ Metzler stable?
What if the eigenvalues of $A$ are real and negative, is $A-B$ Metzler stable?
I believe the answer should be yes. By Gershgorin Circle Theorem, the eigenvalues of $A$ lie within discs centered at the values of $a_{ii}$ with radius equal to to the sum of the non-diagonal entries of row $i$. The matrix $A-B$ will have the Gershgorin discs also centered at $a_{ii}$, but their radius will be smaller. In the limit as the radius goes to zero, the eigenvalues of $A-B$ will tend to be equal to $a_{ii}$, which are negative. However, part of the rightmost disc could lie in the positive half of the plane and I am not sure how to show that the largest eigenvalue of $A-B$ will be negative.
I figured out a way to prove this. There are a few properties of Metzler matrices that we can use to show that this is true:
1) A Metzler matrix $M$ has a real eigenvalue $\lambda_{max}(M)$ such that the real component of any other eigenvalue of $M$ is smaller or equal than $\lambda_{max}(M)$. This is a result of extending the Perron-Frobenius theorem to Metzler matrices.
2) If a Metzler matrix $M_1$ is entry-wise smaller or equal than a Metzler matrix $M_2$, (i.e. $M_1 \preceq M_2$), then $\lambda_{max}(M_1) \leq \lambda_{max}(M_2)$.
3) If $M$ is a stable Metzler matrix, then $DM$ is also a stable Metzler matrix if $D$ is a diagonal matrix with positive diagonal entries (D-Stability).
Now, to show $A-B$ with $a_{ij}-b_{ij}>0$ for all $i,j$ pairs is stable, define a diagonal matrix $D$ with positive entries $$d_{ii} = \max_{j} \frac{a_{ij}-b_{ij}}{a_{ij}}$$ for all $i$. Then,
$$A-B \preceq DA$$ and $$\lambda_{max}(A-B) \leq \lambda_{max}(DA) < 0$$.
Therefore, $A-B$ is a stable Metzler matrix.