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A linear operator $Z$ on a Banach space is called a dissipative operator whenOn a Hilbert space, using the inner product linear in its first argument,If $\mathrm{Re}⟨Zf,f⟩\le 0$, this proves the defining inequality. Conversely that inequality implies $2\lambda \mathrm{Re}⟨Zf,f⟩\le \Vert Zf{\Vert }^2$ for every $\lambda >0$; letting $\lambda \to \infty$ gives

For the next spectral assertion, dissipativity must be retained as a hypothesis. Let $Z$ be closed, densely defined and dissipative. For $\alpha >0$, the dissipative bound makes $\alpha I-Z$ injective. Its range is closed: if $(\alpha I-Z)f_n\to h$, the bound makes $f_n$ Cauchy, and then $Z{f}_n=\alpha f_n-(\alpha I-Z)f_n$ converges. Closedness produces a preimage of $h$. If the range is all of $H$, the same bound gives a bounded inverse, so $\alpha$ is in the resolvent set. If the closed range is proper, there is a nonzero vector in its orthogonal complement, and the definition of the adjoint of a densely defined operator givesThis proves that the positive spectrum of a dissipative operator is adjoint point spectrum:Without dissipativity the PDF's assertion is false. For example, multiplication by $x$ on $L^2(0,2)$ is bounded, closed and densely defined. The value $1$ is spectral: normalized functions supported closer and closer to $x=1$ make $\Vert (Z-I)f\Vert$ tend to zero. But $Z^{\ast }=Z$ has no eigenvector for eigenvalue $1$, since a square-integrable function supported on the singleton $\{1\}$ is zero.

For the given maximal-domain operator $Q$, finite-support sequences belong to $D(Q)$, so the domain is dense. Coordinatewise limits also show it is closed: if $f^{(j)}\to f$ and $Q{f}^{(j)}\to h$ in ${\ell }^2$, each defining coordinate equation passes to the limit, giving $Qf=h\in {\ell }^2$ and hence $f\in D(Q)$.

Put $g_n=2^n{f}_n$ and $h=Qf$. For $n\ge 1$,Since ${\sum }_{n\ge 1}{2}^{-n}|h_n|<\infty$ by Cauchy-Schwarz inequality, $g_n$ has a finite limit ${\ell }_f$. Its increments are square summable as well. Direct summation givesThe original inner-product series converges because $Qf,f\in {\ell }^2$. Taking $N\to \infty$ therefore givesThus $Q$ is dissipative, including its full maximal domain; the limiting boundary term has not been discarded.

To decide generation, solve $(\alpha I-Q)f=h$ for arbitrary $h\in {\ell }^2$ and fixed $\alpha >0$. The first coordinate gives $g_0=f_0=h_0/(\alpha +1)$, and the rest giveBecause all denominators are at least one,Hence $g_n$ is bounded, so $f_n=2^{-n}{g}_n$ belongs to ${\ell }^2$. The coordinate equations give $Qf=\alpha f-h\in {\ell }^2$, proving $f\in D(Q)$ and surjectivity of $\alpha I-Q$. Together with density, closedness and dissipativity, the Lumer-Phillips theorem proves the contraction generator from an exponentially weighted difference operator conclusion:

The domain is crucial. The formal transposed matrix suggests a positive-eigenvalue solutionwhich is square summable because the products converge. It is not a true adjoint eigenvector: the admissible test vector $f_n=2^{-n}$ has $Qf=(-1,0,0,\dots )$, so $⟨Qf,u⟩=-\overline{u_0}$, while $\alpha ⟨f,u⟩=\alpha \overline{u_0}{\sum }_n{4}^{-n}{\prod }_{k<n}(1+\alpha 4^{-k})$. These cannot agree for $\alpha >0$ unless $u_0=0$. This exemplifies how boundary terms exclude formal adjoint eigenvectors, and explains why a calculation using only finite-support test vectors would give the wrong generation answer.

At $(\sigma ,\mu )=(1,0)$ the origin's matrix is $\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)$: it has a double zero eigenvalue and one eigenvector. Trace and determinant vary independently with $\mu$ and $\sigma$ there, so two parameters are needed. The symmetry of an odd function removes quadratic terms, giving a reflection-symmetric cubic double-zero normal form. The quadratic nonlinearity required for a generic Bogdanov–Takens bifurcation vanishes.

Use a prime for differentiation with respect to the rescaled time. Direct substitution, with all rescaled variables written without tildes, givesAt $\epsilon =0$, the equations are $u^{\prime }=-sv+v^3/2$, $v^{\prime }=2u$. Fordifferentiation gives $H^{\prime }=2u(-sv+v^3/2)+(sv-v^3/2)2u=0$. This is the Hamiltonian blow-up of a reflection-symmetric double-zero point.

For $s>0$, the equilibrium points are a centre equilibrium at $(0,0)$ and saddles at $(0,\pm \sqrt{2s})$. The centre equilibrium has eigenvalues $\pm i\sqrt{2s}$; at either saddle equilibrium they are $\pm 2\sqrt{s}$. Levels $0<H<s^2/2$ contain closed inner ovals around the centre equilibrium. The saddle equilibrium level hasThese two branches form a heteroclinic cycle connecting the saddles. The branch with $u>0$ runs from the lower saddle equilibrium to the upper one because $v^{\prime }=2u>0$; the other runs back. Outside the potential well, trajectories are open. The right panel of the preceding figure uses the coordinates $(v,u)$ to display the well and these directed connections.

Assume the consistency regularity implicit in the requested rate, for example $u\in C^4$ on the closed square. Bounded fourth derivatives and Taylor's theorem give a uniform residual ${\tau }_{ij}={\Delta }_{h}u(ih,jh)-f(ih,jh)$ with $|{\tau }_{ij}|\le C_0{h}^2$. Subtracting the exact grid values from the numerical equations gives ${\Delta }_{h}e=-\tau$, with zero boundary error.

Let $A=-h^2{\Delta }_h$ be the unscaled positive Dirichlet matrix. Its eigenvectors are products of discrete sine vectors, which form an orthogonal basis; direct substitution gives eigenvaluesSince $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8h^2$. Therefore $\Vert A^{-1}{\Vert }_2\le 1/(8h^2)$. The error equation is $Ae=h^2\tau$, and its residual has norm $\Vert \tau {\Vert }_2\le M{C}_0{h}^2\le C_{0}h$. Combining the bounds yieldsThis is the unweighted grid error for the five-point Poisson formula. The unweighted norm loses one power of $h$ because the grid has order $h^{-2}$ points; the area-weighted norm is second-order accurate. A regularity assumption is needed for the uniform truncation estimate and hence for this rate; the question does not specify it explicitly.

Choose a nonzero vector in an irreducible representation. Its finite group orbit spans a finite-dimensional invariant subspace, which must be the whole representation. The representing matrices of an abelian group commute. Choose an eigenspace of one matrix; all other matrices preserve it. Repeatedly choosing an eigenspace for the restrictions of the remaining matrices gives a common eigenvector. Its line is invariant under the whole group. Irreducibility therefore makes the representation that line, proving every irreducible complex representation of a finite abelian group is one-dimensional.

A bounded operator on a complex Hilbert space is self-adjoint when $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$. The spectral theorem for compact self-adjoint operators says its nonzero spectrum consists of real eigenvalues of finite multiplicity, with at most countably many and with accumulation possible only at zero. Their eigenspaces are mutually orthogonal; their closed span is the orthogonal complement of the kernel. On that span the operator has the norm-convergent spectral action $Tx={\sum }_j{\lambda }_j⟨x,e_j⟩e_j$.

Take $H={\ell }^2(\mathbb{N})$ and $T{e}_n=e_n/n$. It is self-adjoint and has infinite-dimensional range. Truncation to the first $N$ coordinates is a finite-rank operator and differs from $T$ in norm by $1/(N+1)$, so $T$ is compact.

The resolvent set comprises those $\lambda$ for which $T-\lambda I$ has a bounded everywhere-defined inverse; its complement is the spectrum, and the point spectrum consists of eigenvalues. In this exampleFor $\lambda$ outside the displayed closed set the inverse is diagonal with entries $(1/n-\lambda {)}^{-1}$ uniformly bounded, proving membership of the resolvent. Each $1/n$ has eigenvector $e_n$. Zero is not an eigenvalue, but $\Vert T{e}_n\Vert \to 0$ prevents a bounded inverse; also $(1/n{)}_n\in {\ell }^2$ has no preimage since that would be the non-square-summable constant sequence. This justifies every asserted spectral case.

UseIt has independent complex eigenvectors $(1,-i{)}^T$ and $(1,i{)}^T$, with eigenvalues $i$ and $-i$. Taking these as columns gives an invertible complex matrix $P$ with $P^{-1}DP=\mathrm{diag}(i,-i)$, so $D$ is a diagonalizable matrix over $\mathbb{C}$. It cannot be diagonalized over $\mathbb{R}$, since a real diagonal matrix would have real eigenvalues, whereas similarity preserves the characteristic polynomial.

Take the Jordan blockIts only eigenvalue is one. An eigenvector satisfies $(C-I)v=0$, so its second coordinate vanishes; the eigenspace is one-dimensional. A diagonalizable matrix of size two requires a basis of two independent eigenvectors, which this matrix lacks even over $\mathbb{C}$.

Allow an eigenvector $v\ne 0$ to have complex entries, since real matrices can initially have complex eigenvalues. Write $v^{\ast }$ for its conjugate transpose. Reality of $A$ gives $A^{\ast }=A^T$; symmetry then gives $A^{\ast }=A$. Thus $A$ is a Hermitian matrix. The scalar $v^{\ast }Av$ equals its complex conjugate, because $(v^{\ast }Av{)}^{\ast }=v^{\ast }{A}^{\ast }v=v^{\ast }Av$. If $Av=\lambda v$, thensince $v^{\ast }v>0$. This proves that all eigenvalues are real. The equality $A^{\ast }=A^T$ is precisely where reality was used; a complex symmetric matrix need not be Hermitian, as $iI$ illustrates.

If $Au=\lambda u$ and $Av=\mu v$, then $u^{\ast }Av=\mu u^{\ast }v$, but also $u^{\ast }Av=(Au{)}^{\ast }v=\overline{\lambda }{u}^{\ast }v=\lambda u^{\ast }v$. Consequently $(\mu -\lambda )u^{\ast }v=0$. Distinct eigenvalues force $u^{\ast }v=0$. For real eigenvectors this is ordinary Euclidean orthogonality. These arguments establish real symmetric spectral orthogonality without assuming a diagonalization in advance.

A real orthogonal matrix satisfies $P^{T}P=I$, equivalently $P^{-1}=P^T$. If $A$ is symmetric,so orthogonal similarity preserves symmetry. An arbitrary real invertible change of basis need not do so. For example,The final matrix is not symmetric. Orthogonal similarity preserves real symmetry; general similarity does not.

In the given basis, the matrix is $A=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$. Let $\omega =e^{2\pi i/3}$, a primitive cube root of unity. For each $\xi \in \{1,\omega ,{\omega }^2\}$,Thus these are eigenvectors with distinct eigenvalues $1,{\omega }^2,\omega$, respectively, and form an eigenbasis. SetThe columns are mutually orthogonal with squared norm three, so $P^{-1}=P^{\ast }/3$. The diagonalization and characteristic polynomial areScaling $P$ by $1/\sqrt{3}$ gives a unitary diagonalizing matrix.

Let $x_{\ast }=A^{-1}b$ and $e^{(k)}=x^{(k)}-x_{\ast }$. The Richardson iteration givesBy the spectral theorem, a real symmetric positive definite matrix has an orthonormal eigenbasis with positive eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$. The error in the $i$th eigenvector direction is multiplied by $(1-\tau {\lambda }_i{)}^k$. Thus convergence for every initial vector occurs exactly when $|1-\tau {\lambda }_i|<1$ for every $i$, or $0<\tau {\lambda }_i<2$. Since $\rho (A)={\lambda }_{\max }$,Necessity also follows from an initial error in an offending eigenvector direction. At $\tau =0$ it remains unchanged; at $\tau =2/\rho (A)$ the largest-eigenvalue error alternates sign; outside the interval some error does not decay. Hence neither endpoint can be included.

Each endomorphism has three distinct eigenvalues in a three-dimensional complex vector space. Choose a nonzero eigenvector for each eigenvalue; these vectors are linearly independent and hence form an eigenbasis. Both operators are therefore diagonalizable with diagonal matrix $D=\mathrm{diag}(1,2,3)$.

Their characteristic polynomials areIf a polynomial $p$ annihilates either operator, applying it to each eigenvector gives $p(1)=p(2)=p(3)=0$. Thus its minimal polynomial must contain all three factors. Conversely, their product annihilates the diagonal matrix, soFinally, send an ordered eigenbasis of $S$ to the correspondingly ordered eigenbasis of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence $T=CS{C}^{-1}$: the two endomorphisms are conjugate.

Expand the initial wavefunction in the complete orthonormal energy eigenstates, with $c_n=⟨{\psi }_n,\Psi (0)⟩$. The unitary time evolution isOn the two-dimensional subspace spanned by ${\psi }_1,{\psi }_2$, the observable has matrix $\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Its normalized eigenvectors are the symmetric and antisymmetric combinations. A complete orthonormal eigenbasis is thereforeThe eigenvalues are $1,3,0$, with $0$ infinitely degenerate. The first two are nondegenerate, regardless of the distinct energies $E_n$.

The first measurement in quantum mechanics, with outcome $3$, leaves the state ${\varphi }_2$ up to an irrelevant phase. At time $t$ it has becomeIts probability amplitude for outcome $1$ isThe Born rule gives the two-level observable transition probability

Write $p:U(H)\to PU(H)$ for the scalar quotient. Local strongly continuous sections exist: near a fixed projective unitary, fix its phase by requiring one nonzero matrix coefficient, taken relative to a fixed representative, to be positive real. Thus the pullback $E=\{(z,V):p(V)=\rho (z)\}$ is a locally trivial central extension of the circle by the circle. It is compact, since finitely many closed arcs inside such trivializing neighbourhoods cover the compact base and each pullback is a product with the compact fibre.

Any two lifts commute up to a scalar. Their commutator defines a continuous alternating bicharacter $c(z,w)$ on the base: the scalar commutator is independent of phase choices and multiplicative in each argument. If $z$ is an irrational rotation, its powers are dense and $c(z,z^m)=1$ for every integer $m$. Continuity makes $c(z,w)=1$ for every $w$. Such $z$ are themselves dense, so all the lifts commute, and $E$ is abelian.

The tautological representation $R(z,V)=V$ is strongly continuous. Choose a unit vector $\xi$ and average its rank-one orthogonal projection over Haar measure on $E$:The integrand is trace-norm continuous, so $K$ is positive compact with trace one, and commutes with $R(E)$. A positive eigenvalue has a nonzero finite-dimensional eigenspace invariant under $E$. Commuting unitary matrices on that space have a common unit eigenvector $\eta$. Therefore $R(e)\eta =\chi (e)\eta$ for a continuous character $\chi$ of $E$.

On the central fibre, $R(1,\lambda I)=\lambda I$, so $\chi (1,\lambda I)=\lambda$. Consequentlyis independent of that lift. It is a homomorphism and strongly continuous, because the continuous map on $E$ descends through its quotient to the base. It satisfies $p(U_z)=\rho (z)$. This proves circle projective representations lift to unitary representations using an actual central character, rather than assuming a global choice of phases.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.

Fix the quantum enveloping algebra of sl2 conventionThe printed pair of matrices contains an error. On its first basis vector, the printed matrices give $(EF-FE)v_0=ϵ[2{]}^2{v}_0$, whereas the defining commutator gives $ϵ[2]v_0$. Since $q$ is not a root of unity, $[2]\ne 0,1$: $[2]=0$ would give $q^2=-1$, and $[2]=1$ would give $q^2-q+1=0$. Thus the two printed matrices cannot represent this algebra. Keeping the printed $F$ matrix requires exchanging the two nonzero entries of the printed $E$ matrix. We derive that corrected pair rather than assume a classification theorem.

As $q$ is not a root of unity, successive applications of $E$ multiply $K$ eigenvalues by $q^2$. Finiteness of the spectrum therefore makes $E$ a nilpotent operator. Start from a $K$ eigenvector and apply $E$ until the last nonzero vector $v$; then $Ev=0$ and $Kv=\lambda v$, with $\lambda \ne 0$. The defining commutator gives the highest-weight string for quantum sl2For completeness, at $j=1$ this is the commutator applied to $v$. To advance from $j$ to $j+1$, write $E{F}^{j+1}v=F(E{F}^{j}v)+(K-K^{-1})F^{j}v/(q-q^{-1})$, use $K{F}^{j}v=\lambda q^{-2j}{F}^{j}v$, and combine the two coefficients; their sum is the displayed coefficient at $j+1$.

Let $N$ be the first index with $F^{N}v=0$. Such an index exists because nonzero $F^{j}v$ have distinct $K$ eigenvalues $\lambda q^{-2j}$. The linear span of $v,Fv,\dots ,F^{N-1}v$ is stable under $K,E,F$, so simplicity makes it all of $V$. The distinct eigenvalues make these vectors independent; hence $N=3$. Applying the string formula to $F^{3}v=0$ givesAs $[3]\ne 0$ and $F^{2}v\ne 0$, ${\lambda }^2=q^4$, so $\lambda =ϵq^2$ with $ϵ=\pm 1$.

Choose the basis $v_0=v$, $v_1=Fv/[2]$, $v_2=F^{2}v/[2]$. Then $F{v}_0=[2]v_1$, $F{v}_1=v_2$, $E{v}_1=ϵv_0$ and $E{v}_2=ϵ[2]v_1$. The correct matrices in this basis areAlternatively, preserving the printed $E$ instead requires $F_{21}=1$, $F_{32}=[2]$. Both corrected pairs are related by a diagonal change of basis.