Exercise 10.B.1

Proof. Because $T_{}$

C$hasnorealeigenvalues,if$$\lambda$ is an eigenvalue of $T_{}$

C$,then$$\lambda$$isalsoaneigenvalueof$T_C$withequalmultiplicity.Thedeterminantof$T_C$,whichequalsthedeterminantof$T$,isthusaproductoftermsoftheform$$\lambda \overline{\lambda }=|\lambda {|}^2$, which are all greater than $0$, because $0$ is not eigenvalue of $T$. Hence $\det <!--FUNCTION\; APPLICATION-->T>0$. □