I want to prove, if possible, that the eigenvalues of a product of two special real symmetric $n\times n$ matrices are real and positive. I consider:
a) a diagonal matrix $M$ with positive real numbers.
b) a symmetric matrix $K$ with positive real numbers in the diagonal, $K_{i,i}>0$, and negative real numbers elsewhere, $K_{i,j}\leq 0, \ \ j\neq i$ . Additionally, on each row, the sum of the absolute value of the out of diagonal elements is smaller or equal to the diagonal one, i.e.:
$K_{i,i}\geq\sum_{j\neq i} |K_{i,j}|$.
Here is a $3\times 3$ example:
\begin{align}
M=
\begin{bmatrix}
m_1 \ &0 \ &0 \\
0 &m_2 \ &0 \\
0 \ &0 \ &m_3
\end{bmatrix}
\end{align}
and:
\begin{align}
K=
\begin{bmatrix}
k_1+\kappa_1 \ &-\kappa_1 \ &0 \\
-\kappa_1 &\kappa_1+\kappa_2 \ &-\kappa_2 \\
0 \ &-\kappa_2 \ &k_3+\kappa_2
\end{bmatrix}
\end{align}
I study the eigenvalues of $Z=M^{-1}K$:
$M^{-1}K x=\omega^2 x$
I want to prove that $\omega^2\in\mathbb{R}^+$.
What I know so far is that
$Z$ is not symmetric: $(M^{-1}K)^T=K^T(M^{-1})^T=KM^{-1}$, but that the eigenvalues of $Z$ and $Z^T$ are identical.
However I could not find any other result that could help.
This problem arises in the study of $n$ coupled oscillators with $M$ the matrix containing the value of the masses, and $K$ containing the value of the springs; the coupling is accounted by the out of diagonal elements of $K$. On physical ground I expect the eigenvalue $\omega^2$ to be real and positive (real frequency of oscillations).
Best Answer
By the Gershgorin circle theorem the eigenvalues of $K$ are non-negative. So, $M$ is positive definite and $K$ is positive semi-definite. I can at least show that $M ^{-1}K$ has non-negative eigenvalues. For this, note that $M^{-1}K$ is symmetric with respect to the positive definite inner product $[\cdot,\cdot] = \langle M\cdot,\cdot\rangle$. So, it has real eigenvalues and $[M^{-1}Kx,x] = \langle Kx,x\rangle\ge 0$. Hence, its eigenvalues are non-negative.
EDIT: If the diagonal dominance of $K$ is strict, then $K$ is positive definite, too, and so $M^{-1}K$ has only positive eigenvalues.