$A$ is elliptic iff all eigenvalues of $A$ are strictly positive, or all are strictly negative — i.e. the leading-symbol matrix is definite.
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$A$ is hyperbolic iff all eigenvalues are nonzero and there exists an index $j$ such that every other eigenvalue has sign opposite to that of $\lambda_j$ (signature $(1, n-1)$ or $(n-1, 1)$).
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$A$ is parabolic iff there is exactly one zero eigenvalue (at some index $j$) and all other eigenvalues are nonzero and share a common sign (their pairwise products are positive).
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Spectral theorem (real-symmetric form): for a real Hermitian matrix $A$ with the orthogonal matrix $U$ of eigenvectors, $U^{T} A U = \operatorname{diag}(\lambda_i)$.
The matrix of eigenvectors produced by the spectral theorem is invertible (nonzero determinant), as it is orthogonal.
Normal form for a real Hermitian matrix: there is an invertible $M$ with $M^T A M$ a diagonal matrix whose entries are $+1$, $-1$, or $0$ according to the sign of the corresponding eigenvalue. This is the matrix-level statement behind Hadamard's classification of second-order PDEs.
Helper: when all eigenvalues are positive, the sign-diagonal $\operatorname{diag}(\pm 1, 0)$ collapses to the identity matrix.
Helper: when all eigenvalues are negative, the sign-diagonal collapses to $-I$.
Helper: at a nonzero eigenvalue, the sign expression evaluates to either $+1$ or $-1$.
Hadamard's classification of second-order PDEs (Theorem 3.1, matrix form): for any real Hermitian leading-symbol matrix $A$, there exists an invertible change of variables $M$ bringing $A$ into a diagonal sign normal form, and additionally
- in the elliptic case, $M^T A M = \pm I$,
- in the hyperbolic case, $M^T A M$ is a $\pm 1$-diagonal whose entries split in sign with exactly one differing sign,
- in the parabolic case, $M^T A M$ is a diagonal with exactly one $0$ entry and the remaining entries all $+1$ or all $-1$.