Proof that all the roots of a secular equation are real numbers I am studying the book Introduction to Quantum Mechanics, by Linus Pauling and E. Bright Wilson.
In the chapter dedicated to Perturbation Theory (some methods used to get approximate solutions of Schrodinger's equation) the following algebraic equation, called secular equation, is presented:
det$(A-x·B)=0$ (or det$(a_{ij}-b_{ij}·x)=0$)
where $A$ and $B$ are symmetric (or also hermitian) matrices ($a_{ij}=a_{ji}$, and $b_{ij}=b_{ji}$ or also $a_{ij}=\overline{a_{ji}}$ and $b_{ij}=\overline{b_{ji}}$ for all  $i, j$).
In the book is written that all solutions of this secular equation are real numbers. I am interested in proving this statement, but I don't know how to do it. Can you help me please?
 A: Either you are misquoting the authors or the authors are wrong.
The statement as it stands is wrong because every square matrix is the product of two symmetric matrices. Therefore, if you take a non-singular real matrix $C$ that has some non-real eigenvalues and write it as a product of two symmetric matrices $A$ and $B^{-1}$, then some of the roots of the equation $\det(A-xB)=0$, which are eigenvalues of $C$, must be non-real. For a concrete counterexample, consider
$$
C=\pmatrix{11&-4\\ 7&1}=AB^{-1}=\pmatrix{6&5\\ 5&2}\pmatrix{1&1\\ 1&-2},
$$
whose eigenvalues are $6\pm\sqrt{3}i$.
The statement is true, however, if $A$ and $B$ can be simultaneously diagonalised by congruence and the determinant amounts to a non-constant polynomial. This includes in particular the case where at least one of $A$ and $B$ is positive or negative definite. In this case, let $A=PD_1P^\ast$ and $B=PD_2P^\ast$ where $D_1,D_2$ are real diagonal matrices and $P$ is an invertible matrix (that can be chosen to be real if both $A$ and $B$ are real symmetric). The secular equation then reduces to $\det(D_1-xD_2)=0$, whose roots, if any, are clearly real.
Edit. Apparently, in equation (24-14) in the book, the matrix $B$ (denoted by $\Delta$ in the book, if I read it correctly) is defined as the integral of an outer product. Hence it is positive semidefinite. If it is positive definite, then $A$ and $B$ are simultaneously diagonalisable by congruence, because if $UD_1U^\ast$ is a unitary diagonalisation of $B^{-1/2}AB^{-1/2}$, then $A=PD_1P^\ast$ and $B=PP^\ast$ where $P=B^{1/2}U$. Hence all roots of the secular equation are real.
If $B$ is singular and positive semidefinite, the determinant may be a constant and hence the equation may have no solutions (consider $\det(1-x\cdot0)=0$ for instance) or infinitely many solutions (consider e.g. $\det(0-x\cdot0)=0$), and in the latter case, since every real or complex number $x$ is a solution, the authors are, strictly speaking, wrong. However, if the determinant is a non-constant polynomial in $x$, since the roots of a polynomial equation are continuous in the equation's coefficients and $B$ is the limit of a sequence of positive definite matrices, all roots of the secular equation are real.
