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For $x^{\prime }=Lx$ and $E=\Vert x{\Vert }_2^2$, the Hermitian matrix $H=(L+L^{\ast })/2$ gives $\dot{E}=2x^{\ast }Hx$. Thus some initial condition grows immediately iff the largest eigenvalue of $H$ is positive. All trajectories have nonincreasing energy iff $H$ is a negative semidefinite matrix. This differs from requiring negative real parts for the eigenvalues of $L$: a non-normal matrix can satisfy modal decay while allowing transient growth.

For a one-dimensional quantum tunnelling instanton, the prefactor combines nonzero Gaussian fluctuation eigenvalues with the translation zero-mode Jacobian. Here $L_I=-{\partial }_t^2+V^{\prime \prime }(q_I)/m$, $L_0=-{\partial }_t^2+{\omega }^2$, and the prime excludes the zero mode. Common endpoint regularization and reference normalization define the determinant ratio. The resulting $K$ has frequency units, while $K{e}^{-S_I/\hslash }$ is the dilute crossing rate. This determinant/collective-coordinate structure is also discussed in Dunne, Sulejmanpasic and Ünsal's treatment; unequal neighbouring well frequencies require care and need not share the symmetric-double-well result.

A square nonnegative matrix $A$ is irreducible if for each pair $(i,j)$ there is an integer $k\ge 0$ with $(A^k{)}_{ij}>0$. This says that every index can reach every other through positive entries. The Perron–Frobenius theorem then gives a positive leading eigenvector and an algebraically simple eigenvalue. Irreducibility alone permits other eigenvalues of the same modulus: the cyclic permutation matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has eigenvalues $1,-1$.

With Fourier convention $c_n=N^{-1/2}{\sum }_k{e}^{-ikn}{c}_k$, the Transverse-field Ising model has $H=J{\sum }_k{\Psi }_k^{†}[(g-\cos k){\sigma }_z-\sin k{\sigma }_x]{\Psi }_k$. Expanding the spinor retains the constant $-JgN$ and pairs $k$ with $-k$. The physical quasiparticle energy is twice the positive matrix eigenvalue times $J$, because both momentum partners occur in the Nambu sum.

Two Riemannian manifolds are isospectral for a specified differential operator if their eigenvalues, counted with multiplicities and with the same boundary conditions, agree. Riemannian isometries preserve the Laplace-Beltrami operator spectrum, but the converse can fail. The transplantation theorem and the Sunada theorem give systematic constructions.

Two isospectral manifolds remain isospectral after taking a Riemannian product with the same closed factor: eigenvalues add, with multiplicities given by the product Laplacian eigenspace decomposition. For nonisometric closed connected factors $M_1,M_2$, choose $\epsilon <\delta <{\min }_i\mathrm{inj}(M_i)$. Every closed product geodesic of length less than $\delta$ has constant projection to $M_i$, since a nonconstant closed geodesic shorter than the injectivity radius is impossible. The short coordinate geodesics of the small flat torus span exactly its tangent factor at every point. Any product Riemannian isometry must preserve that intrinsically defined smooth distribution and its orthogonal complement, hence restrict to an isometry of the original factors. Thus these products are still nonisometric. Taking $M_i$ to be four-dimensional tori proves the higher-dimensional stabilization statement without assuming a cancellation theorem for arbitrary products.

A Jordan chain of a linear operator $T$ for an eigenvalue $\lambda$ is a list of nonzero vectors satisfying $(T-\lambda I)v_1=0$ and $(T-\lambda I)v_j=v_{j-1}$ for $2\le j\le r$. These vectors are linearly independent: in a vanishing linear combination, applying the largest relevant power of $T-\lambda I$ isolates the last coefficient times $v_1$, then descending induction removes all coefficients. On their span the operator has one Jordan block. This is different from a generalized eigenfunction that solves $(T-\lambda I)u=0$ outside the original function space.

In a Hermitian neutral-kaon mass matrix, independently choosing square-root signs of $z$ and $z^{\ast }$ can interchange the eigenvalue labels. Choose $z=r{e}^{i\phi }$, $a=\sqrt{r}{e}^{i\phi /2}$ and $b=\sqrt{r}{e}^{-i\phi /2}$, so $ab=r>0$, continuously from the desired CP-conserving limit. Then $(a,-b)$ is the lower-mass eigenvector. At a degenerate off-diagonal zero the mass eigenbasis is not fixed by diagonalization alone.

The kernel of the Graph Laplacian consists exactly of vectors constant on each connected component of a graph. Indeed, $Lx=0$ implies $x^{T}Lx=0$, so the Graph Laplacian quadratic form forces equality at the endpoints of every edge; equality then propagates along graph paths. Conversely, these equalities make every entry of $Lx$ zero. Thus the dimension of the kernel, equivalently the multiplicity of the zero eigenvalue, counts the connected components of a graph.

For the disconnected graph with just two edges, $AB$ and $CD$, in order $A,B,C,D$,The two displayed vectors form a basis of the kernel. The other two eigenvectors $(1,-1,0,0{)}^T$ and $(0,0,1,-1{)}^T$ each have eigenvalue two, so the full list of eigenvalues, including multiplicities, is $0,0,2,2$. Separate connected components of a graph allow independent constant levels; a single global constant is only one of these possibilities.

Suppose positive operators $A,B$ have positive eigenvalues at least $\gamma$ and their nullspaces meet only at zero. If $\vartheta$ is their smallest angle between two subspaces, then $A+B\ge 2\gamma {\sin }^2(\vartheta /2)I$. Indeed $A+B\ge \gamma (2I-P-Q)$, while $(P+Q{)}^2\le (1+\cos \vartheta )(P+Q)$ gives $P+Q\le (1+\cos \vartheta )I$. With a common nullspace, apply the same proof on its orthogonal complement to bound the positive spectral gap.

For positive-definite Hermitian matrices $A,B$, Klein's inequality giveswith equality exactly when $A=B$. To prove it, let $a_i,b_j$ be their eigenvalues and $u_i,v_j$ their orthonormal eigenvectors. The weights $w_{ij}=|⟨u_i,v_j⟩{|}^2$ have row and column sums one. The difference between the two sides isby the scalar logarithm inequality $\ln t\le t-1$. Equality requires $a_i=b_j$ whenever $w_{ij}>0$, implying $A=B$. Limits extend the result to positive semidefinite matrices with the appropriate support condition. Applying it to trace-one matrices proves nonnegativity of quantum relative entropy.

The Kronecker sum of square matrices combines their Kronecker products with identity matrices. If $Ax=\lambda x$ and $By=\mu y$, then $x\otimes y$ is an eigenvector with eigenvalue $\lambda +\mu$. Tensor-product spatial grids therefore turn separated coordinate Laplacians into a Kronecker sum. Two Hermitian matrices give a Hermitian Kronecker sum, allowing an energy method without computing every eigenvector.

For a real symmetric matrix $X$ and $1\le k\le n$,Indeed $Y=Q{Q}^T$ is a rank-$k$ orthogonal projection matrix in the fantope, and projecting onto the eigenvectors of the largest eigenvalues attains the support function maximum. Equivalently, the largest matrix trace of a compression to a $k$-dimensional vector subspace is the sum of the largest eigenvalues.

For circular planetary orbits with angular momenta $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$, the inclination matrix has positive off-diagonal entries, zero row sums, and $L_j{B}_{jk}=L_k{B}_{kj}$. Positive diagonal symmetrization of a matrix with $W=\mathrm{diag}(L_j)$ makes $W^{1/2}B{W}^{-1/2}$ a symmetric matrix. Moreover,All eigenvalues are real and nonpositive. The zero eigenvalue is a common tilt of every orbital plane; nonzero modes describe relative nodal precession.

The Graph Laplacian converts local differences along edges into a linear map. Throughout this section, $G=(V,E)$ is a finite undirected graph with no self-loops or repeated edges, and each edge has weight one. In a fixed order of vertices, let $A$ be the adjacency matrix and $D$ the degree matrix. The combinatorial Graph Laplacian isThus the value at a vertex is compared with its neighbours. Every row sums to zero, so $L1=0$: a constant vector produces no differences. This definition and the associated Graph Laplacian quadratic form are developed in Spielman's authored survey.^[ref]

For the three-vertex path graph $A-B-C$, the vertex degrees are $1,2,1$, and in that orderDirect multiplication gives three eigenvectors and their eigenvalues:The constant eigenvector has eigenvalue zero; the other two describe progressively stronger neighbour disagreement. Dividing these eigenvectors by $\sqrt{3},\sqrt{2},\sqrt{6}$, respectively, gives an orthonormal basis. The four-vertex cycle graph is already worked out under circulant matrix, where Fourier diagonalization of a circulant matrix gives its eigenvalues.

Every formula here uses $L=D-A$ without degree normalization. For comparison, Chung's treatment uses the degree-scaled matrix $D^{-1/2}L{D}^{-1/2}$, with a convention for isolated vertices; its eigenvalues must not be substituted into the examples here.^[ref]