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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.

Under the usual smoothness assumptions for WKB approximation, the strict positive lower bound on $g$ excludes turning point. Put $S(x)={\int }_0^x\sqrt{g(s)}ds$. The oscillatory approximation is $g(x{)}^{-1/4}[A\sin (\lambda S(x))+B\cos (\lambda S(x))]$. The zero boundary value at the origin forces $B=0$, and the one at the right end gives the quantization conditionHere $n$ is large and $C$ is an arbitrary normalization constant. Nonzero solutions exist only at the eigenvalues, not for every arbitrary large $\lambda$. The approximation concerns the leading phase and envelope.

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 $\mathbb{C}$.

TakeIts characteristic polynomial is ${\lambda }^2+1$, whose roots are $i$ and $-i$. Thus this real matrix has no real eigenvalues.

TakeThe real eigenvalues are one and two, with eigenlines spanned by $(1,0{)}^T$ and $(1,1{)}^T$ respectively. These independent vectors form a real eigenbasis, so $E$ is a diagonalizable matrix over $\mathbb{R}$. Their inner product is one, and any two nonzero vectors on those eigenlines have nonzero inner product. Rescaling can normalize their lengths but cannot make them orthogonal. Hence no orthonormal eigenbasis exists.

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.

The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.

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.

Sylvester's law of inertia states that a real quadratic form can be brought by an invertible real change of coordinates towith $r$ remaining zero coordinates, and that the integers $(p,q,r)$ are uniquely determined. To prove existence, represent it by a symmetric matrix. The allowed orthogonal diagonalization yields real eigenvalues, and scaling each nonzero eigen-coordinate by the square root of its absolute eigenvalue changes its coefficient to $+1$ or $-1$.

For uniqueness, $p$ is the largest dimension of a subspace on which the form is positive definite. The positive coordinate subspace attains $p$. Any subspace of dimension greater than $p$ intersects the $(q+r)$-dimensional span of the negative and zero coordinates nontrivially, by the dimension formula. A nonzero vector in that intersection has nonpositive quadratic value, contradicting positive definiteness. Similarly, $q$ is the maximal dimension of a negative-definite subspace. These descriptions are independent of coordinates, and $r=\dim V-p-q$ is then fixed. This proves existence and uniqueness, including degenerate forms.

For the final application, nonsingularity gives $p+q=2m$. In diagonal coordinates write $v=(v_{+},v_{-})$, with $Q(v)=|v_{+}{|}^2-|v_{-}{|}^2$. If $v\in U$ projects to zero in the positive coordinates, $Q(v)=0$ forces $v_{-}=0$, so projection $U\to {\mathbb{R}}^p$ is injective. The same argument applies to the negative coordinates. Hence $m\le p$ and $m\le q$, and their sum is $2m$. ThereforeThe inertia signature is $(m,m)$, or signature difference $p-q=0$ if that convention is used. This is the isotropic dimension bound from real inertia.

The Sturm-Liouville operator satisfies the Lagrange identity for a Sturm-Liouville operatorIndeed the potential terms cancel, and differentiating the right side gives the remaining two derivative terms. Integrate over $[0,1]$. The common Dirichlet boundary conditions make the endpoint term zero, while $L(y_0;{\lambda }_0)=0$. ThereforeThis is the necessary solvability condition imposed by the nonzero homogeneous solution; its proof here does not rely on merely quoting the Fredholm alternative.

For the nonlinear problem, chooseThe orthogonality condition then givesWith $y=ϵy_0+{ϵ}^2{y}_1$, where the correction remains bounded as $ϵ\to 0$, the normalization givesDivide by ${ϵ}^3$ and use $\mu =O(1)$ to obtainThe positive quartic integral determines the leading direction of the nonlinear shift from the linear eigenvalue.

The orbital angular momentum operators are $\mathbf{L}=\mathbf{x}\times \mathbf{p}$, soThe canonical commutation relations are $[x_i,x_j]=[p_i,p_j]=0$ and $[x_i,p_j]=i\hslash {\delta }_{ij}$. On a common domain of smooth wavefunctions, the product commutator identity givesContract this with ${ϵ}_{iab}{ϵ}_{jcd}$. The first term contributes $i\hslash ({\delta }_{ij}\mathbf{x}\cdot \mathbf{p}-x_j{p}_i)$ and the second contributes $i\hslash (x_i{p}_j-{\delta }_{ij}\mathbf{x}\cdot \mathbf{p})$. Their sum isThis verifies the angular momentum commutation relations explicitly. Similarly,

In the position representation $\mathbf{p}=-i\hslash \nabla$. For a radial wavefunction, $\nabla {\Psi }_0={\Psi }_0^{\prime }(r)\mathbf{x}/r$, parallel to $\mathbf{x}$, soWrite ${\Phi }_i=h(r)x_i$ with $h(r)={\Psi }_0^{\prime }(r)/r$. The operator $L_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$ annihilates the radial factor. It gives $L_3{x}_1=i\hslash x_2$, $L_3{x}_2=-i\hslash x_1$, and $L_3{x}_3=0$, henceThus these components, when nonzero, are the specified eigenfunctions with eigenvalues $0,+\hslash ,-\hslash$. This is an example of angular momentum of a radial function times a linear polynomial.

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.