matrix inequalities, bound $k$ in $\frac{d}{ds}K(s)+kK(s)\leq 0$

Question

We have $$K(s) = K^{*}(s)>0; \frac{d}{ds}K(s)<0, K(s) \in C^1[0,\tau] $$ where $K(s)$ is self-adjoint matrix. The question is find the upper bound $k_0$ of $k>0$ where $k$ is the biggest number satisfying inequalities for any $s \in [0; \tau]$ $$\frac {d}{ds}K(s)+kK(s)\leq0$$ I tried if dimension of matrix equeales to $1$, it means we have functions. Then $$K(s)e^{ks}+kK(s)e^{ks}\leq0$$ $$\frac{d}{ds}(K(s)e^{ks})\leq0$$ $$K(s)e^{ks}\leq K(0)$$ $$k\leq \frac{1}{s}\log(\frac{K(0)}{K(s)})$$ And $$k_0 =\min_s \frac{1}{s}\log(\frac{K(0)}{K(s)})$$ If $K(s)$ is matrix i stucked, because can't write inequalities like $k\leq$ something(s).