Given matrices $A\in\mathbb R^{n\times n},C\in\mathbb R^{m\times n}$, for all $\epsilon >0$ does there exists some $L\in\mathbb R^{n\times m}$ such that $\rho(A-LC)=\rho(A)+\epsilon$?
where $\rho(A)$ is the spectral radius of the matrix $A$.
The answer in negative when $C=0$, but i think it is positive in the other case.
My attempt is to write $A$ in a Jordan canonical form and $LC$ in the same basis, then deduce the hypothesis, but i don't know if we can write $LC$ in the same basis since there are rank constraints on $C$ i think.
Yes. Since $\lim_{k\to\infty}\rho\left(\frac1kA-C^TC\right)=\rho\left(\lim_{k\to\infty}\frac1kA-C^TC\right)=\rho(-C^TC)=\|C\|_2^2>0$, we have $\lim_{k\to\infty}\rho\left(A-kC^TC\right)=+\infty$. Therefore $\rho\left(A-kC^TC\right)=\rho(A)+\epsilon$ for some $k$, by the continuity of $\rho$ and the intermediate value theorem.