The *separation* between two square matrices $A$ and $B$, often used as a measure of the sensitivity of invariant subspace problems, is defined as
$$
\operatorname{sep}(A,B)=\min_{X\neq 0}\frac{\left\Vert AX-XB\right\Vert}{\left\Vert X\right\Vert}
$$
for a suitable matrix norm (see e.g. the book by Stewart and Sun or Golub and Van Loan, Section 7.2.4).

Intuitively, it measures the distance between the spectra of $A$ and $B$: if they share an eigenvalue, $\operatorname{sep}(A,B)=0$, and if at least two of their eigenvalues are very close then it is small.

Let $S$ be a stable matrix, i.e., $\Re\lambda<0 $ for each of its eigenvalues $\lambda$. Let $U$ be an unstable matrix, i.e., $-U$ is stable. I would like to prove that $$\operatorname{sep}(S,U)<\operatorname{sep}(S,kU)$$ for $k>1$, or at least some weaker result on the lines of "if I take the eigenvalues more far apart than they are, then the separation increases". For instance, $$\operatorname{sep}(S,O)<\operatorname{sep}(S,U),$$ where $O$ is the zero matrix (of size $1\times 1$, or of the same size of $U$, does not matter) would suit my needs. Establishing this result for at least one among Euclidean and Frobenius norm would be fine.

Is there any known result in this direction, to your knowledge?