General solution of a system with distinct eigenvalues

Question

I'm going through Landau's Mechanics, and on the chapter about oscillations of systems with more than one degree of freedom, he goes from the characteristic equation $$\sum_{k}(-\omega^2m_{ik}+k_{ik})A_k=0$$ and he requires the determinant of the coefficients must vanish $$|k_{ik}-\omega^2m_{ik}|=0$$ so far so standard, but then he substitutes each $\omega_\alpha$ back into the sum to find the corresponding coefficients. At this stage he claims that if all eigenvalues $\omega_\alpha$ are distinct, then the coefficients $A_k$ are proportional to the minors of the determinant with $\omega=\omega_\alpha$. How does he arrive at this result? I know there is an answer to this on the physics stack exchange which amounts to "try it and see that it works". So far my guess is that it has something to do with finding the inverse using Cramer's rule, since he later states that this step is diagonalising the $m+k$ matrix, but in that case I think there would be some extra terms from the $(m+k)$ matrix itself, since it gets left multiplied after the diagonalised matrix $(m+k)D(m+k)^{-1}$. Is there a mathematically rigorous method to come to his conclusion?

Answer 1

Your question can be reduced to finding a kernel vector for $M$, $Mx=0$, where $M$ has co-rank 1.

This has as direct consequence that not all minors $\det M_{ik}$ can be zero, as then the co-rank would be two or larger. $M_{ik}$ is here the sub-matrix $M$ obtained by deleting row $i$ and column $k$, crossing at the matrix element $m_{ik}$.

With one such non-zero minor fixed, set $x_l=(-1)^{k+l}\det(M_{il})$. It follows that $\sum m_{jl}x_l$ for $j\ne i$ is the Laplace expansion for the row $i$ of the determinant of the matrix $M$ modified so that instead of row $i$ the row $j$ is repeated, so that the value zero results. If $j=i$, then this combination is just the determinant of $M$, which is zero by assumption.

Thus we have a non-zero solution.