Self-adjoint operator

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

For a bounded self-adjoint operator $L$, a spectral Weyl sequence at $\lambda$ consists of unit vectors $f_n$ with $\Vert (L-\lambda I)f_n\Vert \to 0$. Such a sequence exists exactly at points of the spectrum of a bounded operator. If no sequence exists, the shifted operator is bounded below, with closed range and zero kernel. Self-adjointness then makes its range dense, so it is invertible. This spectral criterion is different from the equidistribution Weyl criterion.

A singular Weyl sequence for a bounded self-adjoint operator $L$ at $\lambda$ is a spectral Weyl sequence that also converges weakly to zero. Such a sequence exists exactly when $\lambda$ is in the essential spectrum of a bounded self-adjoint operator. An infinite-dimensional shifted kernel gives a weakly null orthonormal sequence. If that kernel is finite-dimensional but the shifted range is not closed, choose approximate null vectors in the kernel complement and extract a weakly convergent subsequence; its limit belongs to both kernel and complement, so is zero.

A densely defined linear operator on a complex Hilbert space is self-adjoint when its adjoint operator has exactly the same domain and agrees there. Symmetry only requires $A\subseteq A^{\ast }$ and does not guarantee self-adjointness. A self-adjoint operator is a closed operator with real spectrum. Domains matter for differential operators and unbounded multiplication operators.

Positivity makes $A-z$ bounded below with dense closed range for every $z<0$, excluding negative spectral values. Conversely if the spectrum is nonnegative, $R=(I+A{)}^{-1}$ has spectrum in $[0,1]$ by the resolvent spectral mapping identity. The numerical-range endpoint approximate-eigenvalue theorem gives $0\le R\le I$, and positive-form Cauchy-Schwarz inequality gives $\Vert Rf{\Vert }^2\le ⟨Rf,f⟩$. For $u=Rf$, $⟨Au,u⟩=⟨Rf,f⟩-\Vert Rf{\Vert }^2\ge 0$.

Symmetry gives $\Vert (A-z)u\Vert \ge |\mathrm{Im}z|\Vert u\Vert$ when $z$ is nonreal. Closedness makes the range closed; its orthogonal complement is $\ker (A-\overline{z})=0$, so it is also dense and hence surjective. This proves the resolvent estimate and reality of the spectrum. Equality of the adjoint domains is needed in the range argument.

For a self-adjoint operator $L$, solving $Lu=f$ requires $f$ to be orthogonal to every $v\in \ker L$, because $⟨v,Lu⟩=⟨Lv,u⟩=0$. When the range is closed, this condition is also sufficient: $(\mathrm{ran}L{)}^{\perp }=\ker L$ and closedness gives $\mathrm{ran}L=(\ker L{)}^{\perp }$. In finite dimensions the range is automatically closed. This Fredholm solvability condition determines amplitude equations by projecting perturbative forcing onto critical eigenfunctions.

A self-adjoint operator $A$ is positive definite when $⟨Av,v⟩>0$ for every nonzero vector $v$ in its domain. In finite dimension it is invertible and has positive eigenvalues. In infinite dimension, strict positivity of individual quadratic-form values need not give a bounded inverse; the stronger uniform bound $⟨Av,v⟩\ge c\Vert v{\Vert }^2$ with $c>0$ supplies coercivity.

For maps $\beta :U\to V$ and $\gamma :V\to W$ between finite-dimensional inner-product spaces, with $\mathrm{im}\beta =\ker \gamma$, the operator $\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma$ is positive definite. Its quadratic form is $\Vert {\beta }^{\ast }v{\Vert }^2+\Vert \gamma v{\Vert }^2$, whose zero set is $\mathrm{im}\beta \cap (\mathrm{im}\beta {)}^{\perp }=\{0\}$. This uses the adjoint operator relation $\ker {\beta }^{\ast }=(\mathrm{im}\beta {)}^{\perp }$.

On ${\ell }^2$, the bounded self-adjoint operator $(Lv{)}_j=v_j/j$ is strictly positive. Yet $f_j=1/j$ lies in ${\ell }^2$ and $Lu=f$ would require the non-square-summable vector $u_j=1$. A positive uniform lower bound, rather than strict positivity alone, guarantees a bounded inverse for a bounded self-adjoint operator.

Every self-adjoint operator on a finite-dimensional real inner-product space has an orthonormal basis of eigenvectors and only real eigenvalues.

A Hermitian operator on a $2^n$-dimensional Hilbert space, with $2^n$ distinct eigenvalues all belonging to $\{c/2^n:0\le c<2^n\}$, must occupy this entire grid. This follows from the pigeonhole principle: the set of possible values and the set of distinct eigenvalues have the same cardinality. Its minimum eigenvalue is zero, so the operator is necessarily singular. Distinctness therefore does not imply invertibility; it forces a one-dimensional kernel in this specific complete-grid setting.

An orthonormal eigenbasis of a linear operator is an orthonormal basis consisting of eigenvectors.

On polynomials with weighted inner productthe operatoris self-adjoint because $e^{-x}Lf=(x{e}^{-x}{f}^{\prime }{)}^{\prime }$. Its eigenvalues on polynomials of degree at most $n$ are $0,-1,\dots ,-n$; its eigenvectors are scalar multiples of the Laguerre polynomials.

The generalized Laguerre polynomial $L_n^{(\alpha )}$ is orthogonal on $[0,\infty )$ for the weight $x^{\alpha }{e}^{-x}$ when $\alpha >-1$.

A Rodrigues formula constructs an orthogonal polynomial by differentiating a weight multiplied by a power. For generalized Laguerre polynomials,