# Landau mechanics - Normal modes of oscillation

In Landau's Mechanics book there's a section in which he explains small oscillations in systems with $$s \geq 1$$ degrees of freedom. He writes the kinetic and potential energies as $$T = \sum_{i, k} \frac{1}{2}a_{ik}(q_0)\dot{q}_i\dot{q}_k \hspace{1cm} U = \sum_{i, k} \frac{1}{2}k_{ik}x_ix_k$$ where $$q_0$$ is a stable equilibrium point, so that the matrix $$K = (k_{ik})$$ is positive definite, and $$x = q - q_0$$. Also, he puts $$a_{ik}(q_0) = m_{ik}$$, so that $$T = \sum_{i, k} \frac{1}{2}m_{ik}\dot{q}_i\dot{q}_k$$ Using Lagrange's equations while looking for solutions of the form $$x_k = A_k e^{i\omega t}, \; A_k, \omega \in \mathbb{C}$$, he obtains $$\sum_k (-\omega ^2m_{ik} + k_{ik})A_k = 0$$ which can be rewritten in matrix form as $$(-\omega ^2M + K)A = 0$$. Both $$M$$ and $$K$$ are positive definite and hence invertible, so the last equation is equivalent to $$(M^{-1}K - \omega^2 I)A = 0$$. So, what Landau is looking for is the eigenvalues and eigenvectors of $$M^{-1}K$$. Then he says that, provided the eigenvalues are all different, the components $$A_k$$ of $$A$$ are proportional to the minors of the determinant of $$(M^{-1}K - \omega^2I)$$, with $$\omega^2$$ eigenvalue.

Why is this? Cramer's rule is useless because the matrix is not invertible.

My reasoning is as follows: put $$C = \frac{M^{-1}K}{\omega^2}$$. Then, we have $$CA = A$$. If we consider the determinant of the matrix C whose $$i$$th column is replaced with $$A$$, we get $$D(C^1, \dots, C^{i-1}, A, C^{i+1}, \dots, C^s) = D(C^1, \dots, C^{i-1}, \sum_j A_jC^j, C^{i+1}, \dots, C^s) = A_i D(C)$$ and so $$A_i = \frac{D(C^1, \dots, C^{i-1}, A, C^{i+1}, \dots, C^s)}{D(C)}= \frac{1}{D(C)}\sum_k M_{ik}A_k$$ where the last equality follows form Laplace's expansion of the determinant in the numerator and $$M_{ik}$$ are coefficients that are proportional to the minors of $$C$$.

How do I conclude that the coefficients $$A_k$$ are proportional to the minors?

EDIT: I'm looking for a proof of this fact.

• Does this answer your question? Why are the solution coefficients for a harmonic oscillator proportional to minors of the determinant? Commented Apr 22, 2020 at 10:07
• Not really, I was looking for a proof... I did read that post though. Commented Apr 22, 2020 at 10:22
• Commented Apr 22, 2020 at 16:49
• I don't fully understand how your post ties into my question... Commented Apr 22, 2020 at 21:37
• Commented Apr 25, 2020 at 19:03

For a given matrix N, you are seeking the null vector, so $$\det (N)=0$$. Now the transpose of the cofactor matrix is $$\operatorname{Adj}(N)=C^T,$$ where C, the cofactor matrix of N, has the properly sign-permuted minors in the corresponding entries. The property of the adjugate is that $$N ~\operatorname{Adj}(N)= 1\!\!1 ~ \det (N).$$ But we assumed $$\det (N)=0$$, so the r.h.side of the above vanishes.