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The Jordan normal form theorem says that every complex square matrix is similar to a direct sum of Jordan blocks, uniquely up to their order. For an eigenvalue $\lambda$, if its blocks have sizes $n_1,\dots ,n_r$, then its algebraic multiplicity and geometric multiplicity areand its contribution to the minimal polynomial is $(t-\lambda {)}^{{\max }_j{n}_j}$. Thuswhere $s_{\lambda }$ is the largest $\lambda$-block size.

For the given matrix,Moreover,HenceThe eigenvalue $-1$ therefore has one Jordan block of size two, so

A shift-invariant basis is a common eigenbasis of all cyclic shift operators $U(\gamma )$. For the displayed Fourier states, orthonormality follows from the root-of-unity filter:There are $N$ vectors in this orthonormal set in the $N$-dimensional Hilbert space, so they form a basis. Reindexing the finite sum givesThus every $|{\xi }_{\alpha }⟩$ is simultaneously an eigenvector of every shift, with eigenvalue ${\omega }^{\alpha \gamma }$ for $U(\gamma )$.

Write $x=g^{\eta }$. Applying $D_x$ and then changing the summation variable from $k$ to $\ell =k-\alpha \eta$ givesHence this state is an eigenvector of $D_x$ with eigenvalue $\boxed{{\omega }^{\alpha \beta \eta }}$.

A complex representation of $G$ is a complex vector space $V$ together with a homomorphismIt is faithful when $\ker \rho =\{e\}$, equivalently when $\rho$ is injective. Every finite group has a faithful complex representation: in the regular representation, $G$ permutes the basis $\{e_h:h\in G\}$ byand a group element fixing every basis vector must be the identity.

If $g$ is conjugate to $g^k$, then $\rho (g)$ is similar to $\rho (g^k)=\rho (g{)}^k$. Thus the two matrices have the same spectrum, while an eigenvalue $\lambda$ of $\rho (g)$ gives the eigenvalue ${\lambda }^k$ of $\rho (g{)}^k$. Hence

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\le k<p$. Faithfulness makes $\rho (g)$ have order $p$, so its spectrum contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct valuesin the spectrum. Therefore $\dim V\ge p-1$. This is the spectrum orbit bound for a faithful symmetric-group representation.

Finally let $G$ be the group of all permutations of $\mathbb{N}$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a subgroup $(C_2{)}^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the group of diagonal sign matrices, which has order $2^d$. A faithful restriction would require $2^n\le 2^d$ for every $n$, an impossibility.

Set $c=⟨\psi |\varphi ⟩$. The off-diagonal Hermiticity identity holds automatically, whilemust be real. Thusis necessary and sufficient. For $g\ne 0$, put $r=gc/|g|\in \mathbb{R}$. Since the two normalized spanning vectors are linearly independent, $|r|<1$.

The normalized eigenvectors and eigenvalues areDirect application of $H$ verifies the eigenvalue equations. Their unnormalized inner product isso the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If $g=0$, $H=0$ and every vector is an eigenvector.

At $r=1$, the Jacobian at the origin isIts eigenvalues are $0,-2,-1$. The zero eigenvalue makes the equilibrium nonhyperbolic.

For $\psi ={\sum }_n{\alpha }_n{\varphi }_n$ and an orthonormal trial set,Stationarity subject to ${\alpha }^{†}\alpha =1$ givesThe finite-subspace variational method, equivalently the Rayleigh-Ritz variational principle, says that the minimum quotient is the smallest eigenvalue of $\mathcal{H}$. It is therefore the optimal upper bound on $E_0$ available from this span.

The kinetic matrix is diagonal:For the linear potential,andThus, in units ${\hslash }^2/(m{a}^2)$,The lower eigenvalue gives the two-mode variational bound for a linearly tilted square well

WriteProjecting the equation onto modes $|n|\le D$ gives the Fourier-Galerkin matrix for a drift-diffusion equationwhere ${\hat{w}}_l=0$ for $|l|>d$.

For $w=\cos \pi x$, only
${\hat{w}}_{\pm 1}=1/2$ are nonzero. HenceThe column belonging to the constant mode $m=0$ is zero, soRemove that zero mode. In every remaining row $n$, the sum of the magnitudes of the off-diagonal entries is at most ${\pi }^2|n|$; for $|n|=1$, the would-be coupling to the removed zero mode vanishes. The diagonal entry is $-{\pi }^2{n}^2$. Gershgorin's theorem therefore places every eigenvalue of the nonconstant block inTogether with the zero eigenvalue, all eigenvalues of $B$ have nonpositive real part.

At late times the first sine mode dominates, with growth exponentThusand the population grows precisely when $\mu >{\mu }_c$. Faster diffusion carries organisms to the lethal ends more quickly and raises the threshold; a longer channel lowers the principal diffusion eigenvalue and therefore lowers the threshold.

Using one even and one odd site per unit cell, Bloch theorem reduces the Hamiltonian toIts eigenvalues are

Let $D=\mathrm{diag}(A)$. The Jacobi method isso $H=I-D^{-1}A$ and $v=D^{-1}b$. If $A$ is strictly diagonally dominant, every Gershgorin disc of $H$ is centered at zero and has radiusEvery eigenvalue therefore has modulus below one, so $\rho (H)<1$ and part (i) proves convergence.

This is the unilateral left shift. It is not compact because $S{e}_{n+1}=e_n$. For every $|\lambda |<1$,is an eigenvector with eigenvalue $\lambda$. The norm bound and closedness then give

This is the unilateral right shift. It is not compact because the orthonormal sequence $(T{e}_n)=(e_{n+1})$ has no convergent subsequence. Its norm is one. For $|\lambda |>1$ the Neumann series gives a resolvent; for $|\lambda |<1$, $\bar{\lambda }$ is an eigenvalue of the adjoint left shift, so $\lambda$ lies in the spectrum of $T$. Closedness supplies the boundary. Hence

The action integral can be evaluated exactly:As $\lambda \to \infty$,Its leading logarithmic term is $\lambda \log \lambda$. Hence the large eigenvalues satisfy, to leading order,

If the eigenvalues of a density operator are $p_j\ge 0$ with ${\sum }_j{p}_j=1$, thenwith equality exactly when one eigenvalue is one, namely for a pure state. In dimension $d$, Cauchy--Schwarz gives $\gamma \ge 1/d$, with equality for the maximally mixed state. On an infinite-dimensional space the universal lower bound is zero as an infimum.

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.