Suppose $A \in M_n(\mathbb R)$ is stable. By stable, we mean the eigenvalues are all on the left open half plane of $\mathbb C$. Now if we decrease the value of $A_{11}$, does the matrix remain stable?
I first thought in terms of Gershgorin Disks. If we decrease the entry $A_{11}$, the center of corresponding disk would move to the left of the real axis. But then I realized this is not enough since we only know the eigenvalues are contained in the union of all disks. However, I could not see a counterexample.
Alternatively, the question is a perturbation with rank-one matrix, i.e., we want to know whether $A-t e_1e_1^T$ remains stable for $t > 0$ where $e_1 = (1, 0, \dots, 0)^T$.
Let $A_t=A-te_1{e_1}^T$ and $\lambda_t$ be the max of the real parts of the eigenvalues of $A_t$.
By continuity of $\lambda_t$ wrt ${A_t}_{1,1}$, $A_t$ remains stable when $t$ is small enough.
Yet, there is no lower bound for the admissible variation of $t$, as shows the folllowing example.
Let $\epsilon$ be a fixed small positive real (for example $\epsilon<0.1$) and
$A=\begin{pmatrix}-2-\epsilon&2\\-1&1-\epsilon\end{pmatrix}$, that is, $A_t=\begin{pmatrix}-2-\epsilon-t&2\\-1&1-\epsilon\end{pmatrix}$. Note that $spectrum(A)=-\epsilon,-1-\epsilon$.
The derivative in $A$ of the greatest eigenvalue of $A$ wrt $A_{1,1}$ is $\lambda '=-1$. Then $\lambda_{\epsilon}\approx -\epsilon+\epsilon \approx 0$.
In particular, $A_{2\epsilon}$ is not stable.