$Q(\lambda)x=0$ and $Q(\lambda) = \lambda^2 M+\lambda C +K$ are defined in this PDF file
The matrices $M$, $C$, $K$ are $n\times n $ matrices. The thesis said that when $M$, $C$, $K$ are real or hermitian. the eigenvalues are real or come in pairs $(\lambda,\overline{\lambda})$. But the explanation in the thesis is hard to understand. Would you give me a more clear explanation?
If a square matrix has real entries, its characteristic polynomial has real coefficients as well. Therefore its non real roots must come in pairs $\lambda$ and $\bar{\lambda}$, because if $\lambda\in\mathbb{C}$ is a root of a real polynomial $f(x)$, then $f(\lambda)=0$ and so $$ 0=\bar{0}=\overline{f(\lambda)}=f(\bar{\lambda}) $$ and therefore also $\bar{\lambda}$ is a root of $f$.
For a real symmetric matrix every eigenvalue is real. Suppose $M$ is real symmetric and $\lambda$ is an eigenvalue, with eigenvector $v\ne0$ (possibly with complex entries). Then, denoting by $^H$ the Hermitian transpose, $$ \lambda v^Hv=v^H(\lambda v)=v^HMv=v^HM^Hv=(Mv)^Hv=(\lambda v)^Hv=\bar{\lambda}v^Hv $$ Therefore $(\lambda-\bar{\lambda})(v^Hv)=0$. Since $v\ne0$, also $v^Hv\ne0$ and therefore $\lambda-\bar{\lambda}=0$, which means $\lambda$ is real.