= Solution
A real symmetric $3\times3$ <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 $\hat p_k\hat p_l$:
$$
B_{ij}=\frac1\rho\big[(\lambda+\mu)\hat p_i\hat p_j+\mu\delta_{ij}\big].
$$
Using
$$
\alpha^2=\frac{\kappa+4\mu/3}{\rho}=\frac{\lambda+2\mu}{\rho},\qquad\beta^2=\frac\mu\rho,
$$
gives
$$
\boxed{B_{ij}=(\alpha^2-\beta^2)\hat p_i\hat p_j+\beta^2\delta_{ij}.}
$$
For $\mathbf a\parallel\hat{\mathbf p}$ 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>. \b[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.
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