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A real symmetric $3\times 3$ matrix has three real eigenvalues and a complete orthonormal basis of eigenvectors. For a stable elastic material, strong ellipticity makes ${\mathbf{a}}^{\mathsf{T}}B\mathbf{a}>0$ for every nonzero polarization, so its eigenvalues are positive squared wave speeds. There are therefore three orthogonal plane-wave polarizations in a given direction, with three potentially distinct speeds in an anisotropic medium. Degenerate eigenvalues allow arbitrary orthonormal combinations within their eigenspace; they do not remove polarization degrees of freedom. General anisotropic modes need not be exactly longitudinal or transverse.

For an isotropic medium let $\lambda =\kappa -2\mu /3$. Contract the isotropic stiffness with ${\hat{p}}_k{\hat{p}}_l$:UsinggivesFor $\mathbf{a}\parallel \hat{\mathbf{p}}$ the longitudinal projector has eigenvalue one, so $B\mathbf{a}={\alpha }^2\mathbf{a}$: this is the P wave. For either of two independent $\mathbf{a}\perp \hat{\mathbf{p}}$ the projector vanishes, so $B\mathbf{a}={\beta }^2\mathbf{a}$: these are two degenerate transverse S wave polarizations. An isotropic solid has one P speed and one S speed, but three polarization modes: one longitudinal and two transverse. A fluid limit with zero shear modulus has no propagating S wave, so the positive-speed conclusion requires a solid with positive shear stiffness.

Strong ellipticity requires the displayed inequality for every nonzero displacement vector $m$ and material direction $\nu$. Equivalently, every acoustic tensor is positive-definite. It is weaker than positivity of the moduli on all matrices. A quadratic minor null Lagrangian vanishes on rank-one matrices, so adding such a term does not change the acoustic tensor or this condition.