= Solution
Choose <generalized coordinates> $q_i$ measured from a static <equilibrium point>, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the <small oscillation> <Lagrangian>
$$
\boxed{\mathcal L=\frac12\dot{\mathbf q}^{\,T}A\dot{\mathbf q}-\frac12\mathbf q^TB\mathbf q}.
$$
Here $A$ is the positive-definite symmetric <mass> <matrix> evaluated at equilibrium, and $B$ is the symmetric <Hessian matrix> of the <potential energy> there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic <matrix> contribute only at higher order.
The <Euler-Lagrange equations> are $A\ddot{\mathbf q}+B\mathbf q=0$. For a <normal mode> $\mathbf q=\mathbf v e^{i\omega t}$, they give the <generalized eigenvalue problem> $(B-\omega^2A)\mathbf v=0$. A nonzero <mode shape> exists exactly when
$$
\boxed{\det(B-\omega^2A)=0}.
$$
This is a degree-$n$ <polynomial> in $\omega^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to <zero modes>.
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