While studying normal modes for a system of n coupled oscillators, I have to show that the matrix responsible for motion has real eigenvalues. The matrix has the form $M^{-1}K$, where $M$ is a real diagonal matrix with positive entries and $K$ is a real symmetric matrix.
To show this, the book shows a preliminary result:
Let $A_1,A_2$ be two distinct eigenvectors (linearly independent) with the same eigenvalue $\lambda$ such that $A_1^TMA_2\neq0$, then one can always construct an eigenvector $A_2'$ such that $A_1^TMA_2' = 0$.
This result can be extended to construct a basis $A_\alpha$ for the eigenspace of $\lambda$ such that $A_\alpha^T MA_\beta = 0$ for all $\alpha\neq\beta$. (This is a very similar result to Gram-Schmidt).
Now let $A$ be an eigenvector of $M^{-1}K$ with eigenvalues $\lambda$. This means that the complex conjugate $A^*$ is also an eigenvector with eigenvalue $\lambda^*$:
$M^{-1}KA = \lambda A \Rightarrow M^{-1}KA^* = \lambda^* A^*$
In the book it says that by using the precedent result we conclude that $(A^*)^TMA = 0$. This then implies that since $M$ is diagonal and has real positive entries than $A=0$ or that $\lambda = \lambda^*$, thus proving what we were looking for.
This being a physics book I assume that certain results be obtained with not much rigor, but in this instance I really can't see how the preliminary result implies $(A^*)^TMA = 0$, and thus the desired result.
This begs then the question: how can we prove that $M^{-1}K$ has real eigenvalues?
Based on past experience with proving similar results I assume that $(A^*)^TMA=0$ is the result we aim for (and it this quite natural, since it says that two eigenvectors corresponding to two difference eigenvalues are orthonormal with respect to this scalar product), but I can't find the way to prove this. Anyone can help? Thanks.
Best Answer
$M^{-1}K$ has the same eigenvalues as the similar matrix $M^{-1/2}KM^{-1/2}=M^{1/2}(M^{-1}K)M^{-1/2}$. One can calculate $$\det(\lambda I-M^{1/2}(M^{-1}K)M^{-1/2})=\det(M^{1/2}(\lambda I- M^{-1}K)M^{-1/2})=\det(\lambda I-M^{-1}K)$$ and hence $M^{1/2}(M^{-1}K)M^{-1/2}$ and $M^{-1}K$ have the same characteristic polynomial and therefore the same eigenvalues. Now $M^{-1/2}KM^{-1/2}$ is real symmetric since $K$ is real symmetric and $M^{-1/2}$ is real diagonal. Therefore $M^{-1/2}KM^{-1/2}$ and hence $M^{-1}K$ have real eigenvalues. To be precise, if $d_1,...,d_n$ are the positive diagonal entries of $M$, then $M^{-1/2}$ is defined as the diagonal matrix with the positve entries $d_1^{-1/2},...d_n^{-1/2}$.