# Lagrangian mechanics - small oscillations around equilibrium diagonalization

In my analytical mechanics class, we have been taught that normal modes of small oscillations around equilibrium are given by the solution of $$p(\omega) = \det(K-\omega^2M) = 0$$ Where $K_{ij} = -\frac{\partial^2\mathscr{L}}{\partial q_i\partial q_j}$ and $M_{ij} = \frac{\partial^2\mathscr{L}}{\partial\dot{q}_i\partial\dot{q}_j}$, where $\mathscr{L}$ is the Lagrangian of the problem, and is the form $\mathscr{L}\left({\{}q_i{\}},{\{}\dot{q}_i{\}}\right) = T\left({\{}\dot{q}_i{\}}\right) - V\left({\{}q_i{\}}\right)$, where $i=1,...,n$, $n$ number of generalized coordinates, $T,V$ are kinetic and potential energy respectively.

We were told that $p$ will always have $n$ solutions for $\omega^2$ (could be degenerate), which is due to the fact that $M^{-1}$ is invertible, and $M^{-1}K$ is always diagonalizable.

Is this true? Why is $M^{-1}K$ always diagonalizable (in this particular Lagrangian)?

• Observe that $\det(K-\omega^2M) = \det(M)\det(M^{-1}K-\omega^2I)=\det(M)\det(M^{-\frac12}KM^{-\frac12}-\omega^2I)$. Now $M^{-\frac12}KM^{-\frac12}$ is diagonalisable and $\det(M)\neq0$ by hypothesis. However $M^{-1}K$ is not guaranteed to be diagonalisable unless, e.g., $K$ and $M$ commute. Perhaps there are extra assumptions on the Lagrangian? Jan 18 '16 at 17:04
• Suggestion to the question (v2): Replace the sentence $M^{-1}$ is invertible with $M$ is invertible. Jan 18 '16 at 19:17
• Related: physics.stackexchange.com/q/78500/2451 and links therein. Jan 18 '16 at 19:22

The answers to both your questions basically boil down to "because the kinetic energy is only zero when the system is at rest, and is positive otherwise." However, the answer to the second question in particular gets pretty deep into the linear-algebraic weeds, so fasten your seatbelt.

1. Why is $M$ invertible?

This is basically an assumption about the form taken by the kinetic energy. Roughly speaking, it will generally be the case that if any of our coordinates is changing, then there will be some kinetic energy associated with this change. If we think of the kinetic energy $T(\dot{q}_i)$ as a function of the velocities, we do not expect that there will be any "flat" directions in velocity-space; if there were, this would mean that we could set the system in motion in such a way that there would be no kinetic energy associated with this motion.

Mathematically, this statement is enforced by the idea that $M$ is non-degenerate. A matrix $M_{ij}$ is degenerate if there exists a non-zero vector $v_i$ for which $\sum_j M_{ij} v_j = 0$. This means, in particular, that $M$ has a non-trivial kernel, and thus that it is not invertible. On the other hand, if such a vector exists, then $\sum_{i,j} M_{ij} v_i v_j = 0$. If we think of $v_j$ as representing a column vector of velocities, this would imply that there's a "flat direction" in velocity space that we don't want: there is some combination of velocities that results in zero kinetic energy (since $T \approx \frac{1}{2} \sum_{i,j} M_{ij} v_i v_j$). Therefore, if there are no "flat" directions in velocity space, we must have a non-degenerate $M$, which means it is invertible.

2. Why is $M^{-1}K$ diagonalizable?

While we only needed non-degeneracy for the above part $M$ usually obeys a slightly stronger condition than being non-degenerate; it is usually positive definite, which translated to physical terms means that the kinetic energy is always positive unless the system is at rest, in which case it's zero. Mathematically, this translates to the following condition on $M$: for all vectors $v$, $\sum_{i,j} M_{ij} v_i v_j \geq 0$, with equality only when $v_i = 0$.

Since $M$ is a symmetric matrix, it can be diagonalized; in other words, there exists an orthogonal matrix ($R^T = R^{-1}$) such that $$\tilde{M} = R M R^T$$ is a diagonal matrix whose entries are the eigenvalues of $M$. Moreover—and this is where the positive-definiteness comes in—all of these entries are positive. (A symmetric positive definite matrix cannot have a non-positive eigenvalue, since this would imply that $M_{ij} w_i w_j \leq 0$ for the corresponding eigenvector $w_i$.) This means there is another diagonal matrix $S$ which satisfies $$S^2 = \tilde{M}^{-1}.$$ (The entries of $S$ are just $1/\sqrt{\mu_i}$, where the $\mu_i$ are the diagonal entries of $\tilde{M}$, aka the eigenvalues of $M$.) In particular, this means that $$M^{-1} = R^T \tilde{M}^{-1} R = R^T S^2 R.$$

So, with this in mind: let's suppose we want to solve the eigenvalue problem $$M^{-1} K \vec{v} = \lambda \vec{v}.$$ Define $\vec{v}'$ such that $\vec{v} = R^T S \vec{v}'$. Then the above equation becomes $$(R^T S^2 R) K R^T S \vec{v}' = \lambda R^T S \vec{v}'$$ Left-multiplying both sides by the inverse of $R^T S$, we get $$S R K R^T S \vec{v}' = \lambda \vec{v}'.$$ But the matrix on the left-hand side is symmetric: $(S R K R^T S)^T = S R K R^T S$. In other words, this is now have a conventional eigenvalue problem, and there will be $n$ eigenvectors and eigenvalues that form a basis for the vector space.

Note, importantly, that the eigenvectors $\vec{v}$ will not be orthogonal, as one might expect. This is because we changed basis via a non-orthogonal transformation ($S$ is not an orthogonal matrix), and vectors that are orthogonal before such a transformation will not necessarily be orthogonal afterwards. The eigenvectors $\vec{v}'$ are guaranteed to be orthogonal by the properties of real symmetric matrices, but the corresponding vectors $\vec{v} = R^T S \vec{v}'$ will not be.

• Perhaps it could be interesting to remark that the procedure simultaneously diagonalizes two quadratic forms and not two operators, which would be impossible if they do not commute as in the present case, in general... Jan 18 '16 at 18:19
• Thank you very much. And $n$ eigenvectors $\vec{v}$ is due to the fact that $R^TS$ is invertible and particularly one-to-one, right? Jan 18 '16 at 18:21