= Solution
For a nonzero, time-independent <vector> $\hat{\mathbf u}$, substitute $\mathbf u(t)=\hat{\mathbf u}e^{i\omega t}$. Taking two time <derivatives> gives
$$
\ddot{\mathbf u}=(i\omega)^2\hat{\mathbf u}e^{i\omega t}=-\omega^2\hat{\mathbf u}e^{i\omega t}.
$$
Cancel the nonzero exponential to obtain the <eigenvalue> problem
$$
\boxed{A\hat{\mathbf u}=\lambda\hat{\mathbf u},\qquad \lambda=-\omega^2,\qquad \hat{\mathbf u}\ne0.}
$$
Thus an <eigenvector> specifies a <normal mode> shape and its negative <eigenvalue> specifies the square of its <angular frequency>, $\omega=\sqrt{-\lambda}$. Real <displacements> are obtained by taking the <real part>, or by combining the two temporal solutions as $\hat{\mathbf u}(a\cos\omega t+b\sin\omega t)$.
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