Exercise 10.B.2

Proof. By the previous exercise, $T$ has at least one eigenvalue $\alpha$, which is negative because $\det <!--FUNCTION\; APPLICATION-->T<0$. Suppose by contradiction that $\alpha$ is the only eigenvalue of $T$. Then this can’t be the only eigenvalue of $T_{}$

C$,otherwisethe$$detT$ would equal ${\lambda }^{\dim <!--FUNCTION\; APPLICATION-->V}$ which is positive (because $\dim <!--FUNCTION\; APPLICATION-->V$ is even). The other eigenvalues of $T_{}$

C$arenonreal,sotheycomeinpairsandwo{n}^{\prime }tchangesignof$$detT$. It follows that $\det <!--FUNCTION\; APPLICATION-->T$ is a product of absolute values times $\lambda$ raised to its multiplicity. Thus the multiplicity of $\lambda$ must be odd. This is a contradiction, because the sum of the multiplicities of the other eigenvalues of $T_{}$

C$iseven(becausetheycomeinpairs)andthesumofallmultipliciesmustequal$$dimV$, which is even. Hence, our assumption that $\lambda$ was the only eigenvalue of $T$ is false. □