Specral Radius stability condition A Multi-Input Multi-Output system with stable transfer matrix $L(s)$ is considered. The theorem states that if $\rho(L(j\omega)<1$ then the closed-loop system is stable. In place of the actual theorem its reverse is proved: if the closed-loop is unstable (that is the Nyquist plot of $\det(I+L(s))$ does encircle the origin) then there exists an eigenvalue $\lambda_i(L(j\omega))$ which is larger than $1$ at some $\omega$. The textbook says: if $\det(I+L(s))$ does encircle the origin then there must exists a gain $\epsilon \in (0, 1]$ and a frequency $\omega'$ such that: $$\det(I+\epsilon L(j\omega'))=0$$ I really cannot figure out the reason why this holds.

This is the standard homotopy argument. The closed curve $$ \gamma_\epsilon(\omega)=\det(I+\epsilon L(j\omega)) $$ depends _continuously_ on $\epsilon$ and $\omega$ (the property of determinant). We know that for $\epsilon=1$ it encircles the origin at least once, and for $\epsilon=0$ it reduces to the point $(1,0)$ which does not encircle the origin. Therefore, due to continuity, the origin must escape from the area inside the curve to the outside at some $\epsilon'\in(0,1)$, that is for this $\epsilon'$ the origin must lie exactly on the curve $$\gamma_{\epsilon'}(\omega')=0.$$