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The absolute value of a linear operator $L$ is the positive square root of an operator $L^{†}L$. Its eigenvalues are the singular values of $L$, its trace is the trace norm of $L$, and it appears in the polar decomposition of a bounded operator $L=U|L|$.

On a complex inner product space $V$, with the inner product conjugate-linear in its first argument, an everywhere-defined linear map $T:V\to V$ admits an adjoint on $V$ if there is a linear map $T^{\ast }:V\to V$ satisfyingThe adjoint, when it exists, is unique: the difference of two candidate values at $y$ is orthogonal to every vector and hence to itself. In a finite-dimensional vector space, it always exists, and its matrix in an orthonormal basis is the conjugate transpose. In a space that is not complete, existence must instead be checked.

The identity $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y\in V$ supplies such an adjoint directly, namely $T^{\ast }=T$. This algebraic statement should be distinguished from self-adjointness of a linear operator on a completed Hilbert space, which additionally compares the domains of an operator and its Hilbert-space adjoint. The Gaussian polynomial operator is a useful example where the identity holds on the entire polynomial inner product space and can be used on every finite-dimensional invariant polynomial subspace.

Suppose a family of linear operators, including its limiting operator, is pointwise dominated by a sublinear operator of weak type (1,1). Convergence almost everywhere on an $L^1$-dense subspace then extends to every $L^1$ input. Approximate the input by errors of norm at most $2^{-2n}$. The weak estimate and the Borel-Cantelli first lemma make the corresponding domination errors tend to zero almost everywhere. Linearity bounds the difference between the full and approximating convergence errors by twice that domination error. This argument does not need an uncountable supremum to be measurable.

An annihilating polynomial of a linear operator is a nonzero polynomial $p$ for which the displayed operator is zero. The minimal polynomial is its unique monic annihilating polynomial of least degree and divides every other annihilating polynomial. For a diagonalizable operator with distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_s$, the product ${\prod }_j(t-{\lambda }_j)$ annihilates it. For an individual vector it suffices to annihilate the eigenvalues contributing to that vector; this gives exact finite-step termination bounds for a Krylov subspace method.

A bounded linear operator between normed spaces is approximable if it is an operator norm limit of bounded finite-rank operators. If the target is a Banach space, every approximable operator is a compact operator. For a given tolerance, approximate the operator in norm by one finite-rank operator, cover that operator's bounded unit-ball image by finitely many small balls in its finite-dimensional range, and transfer the cover across the operator-norm error. This makes the original image totally bounded; its closure is complete in the Banach target and therefore compact. The converse on arbitrary Banach spaces would require an additional approximation hypothesis and is not part of this assertion.

A Bogoliubov vacuum is annihilated by the quasiparticle linear operators that diagonalize a stable quadratic bosonic Hamiltonian. If $a_k=u_k{b}_k-v_k{b}_{-k}^{†}$, its original-mode occupation number is $|v_k{|}^2$. It is generally a multimode squeezed vacuum rather than the original Fock vacuum. In infinitely many modes, existence in the original Fock space requires a separate convergence analysis.

If $D=[A,B]$ commutes with $A$, then $[A^n,B]=nD{A}^{n-1}$. A convergent power series therefore gives $[F(A),B]=D{F}^{\prime }(A)$ on a common domain where termwise operations are justified. For unbounded linear operators, the domain condition is essential.

For finite-dimensional complex linear operators with $C=[A,B]$ commuting with both $A$ and $B$, differentiating $e^{-vB}A{e}^{vB}$ gives the constant derivative $C$. Hence $e^{-B}A{e}^B=A+C$, and conjugating the matrix exponential gives $e^{-B}{e}^{-A}{e}^B=e^{-A-C}$. Since $[A,C]=0$, multiplication by $e^A$ proves the identity. The commutator order determines the minus sign.

Fix the convention: the columns of the change-of-basis matrix $P$ are the new basis vectors written in the old basis. Thus $P$ converts new coordinate vectors into old ones; its matrix inverse converts old coordinate vectors into new ones. For the preceding basis,Its determinant is $-2$, so $P$ is an invertible matrix. Checking which direction $P$ converts is essential: different books may call either $P$ or $P^{-1}$ the change-of-basis matrix.

For a linear operator, convert the input into the old basis, apply the old matrix, and convert the output back. Reading that sequence from right to left givesFor example, let the old diagonal matrix stretch horizontal components by $2$ and vertical components by $3$. ThenThe output represents $2u+2v=(4,0)$, just as $A_{\mathrm{old}}(2,0{)}^T=(4,0{)}^T$. The matrices look different because their bases differ, but they encode the same linear operator. This relation is called matrix similarity; it preserves the underlying action.^[ref]

For bounded linear operators, or formally in an associative algebra,Differentiate the two exponentials using the product rule: the result is $e^{\lambda A}[A,B]e^{-\lambda A}$. Induction gives the $n$th derivative $e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$, and the Taylor series at zero proves the formula. For bounded operators convergence follows from $\Vert {\mathrm{ad}}_A^n(B)\Vert \le (2\Vert A\Vert {)}^n\Vert B\Vert$. For unbounded quantum operators, use a common invariant domain and justify the series there, or use the finite-dimensional system of commutators when their span closes.

For bounded linear operators, define $C(\beta )=[A,e^{-\beta B}]$. The product rule gives $C^{\prime }+BC=[B,A]e^{-\beta B}$ and $C(0)=0$. Multiplying by the integrating factor $e^{\beta B}$ yieldsThe identity is useful when the two linear operators do not commute: it preserves their order rather than incorrectly differentiating an operator exponential as if all its factors were scalars. For unbounded linear operators, the products and differentiations need a common justified operator domain.

For linear operators on a common invariant domain, $[A,BC]=[A,B]C+B[A,C]$. The identity follows by adding and subtracting $BAC$ in the commutator.

Suppose linear operators $A,B$ obey $[A,B]=cB$, where $c$ is a scalar, on a common invariant operator domain. The commutator product rule givesFor $n=0$ both sides vanish. This identity explains how powers of a creation operator shift the number operator eigenvalue.

Two linear operators $A$ and $B$ commute when $AB=BA$. Their matrix exponentials then satisfy $e^{A+B}=e^A{e}^B=e^B{e}^A$.

Fix paired orthonormal bases and let $|\varphi {⟩}_A={\sum }_j{c}_j|j{⟩}_A$. Its conjugate index vector in the other Hilbert space uses the conjugated coefficients. For the unnormalized maximally entangled state $|\tilde{\Psi }⟩={\sum }_j|j{⟩}_A|j{⟩}_B$, the partial inner product is $(I\otimes ⟨{\varphi }_B^{\ast }|)|\tilde{\Psi }⟩=|\varphi {⟩}_A$. Both the conjugation and the lack of normalization of $\tilde{\Psi }$ matter. Applying a linear operator on the first subsystem gives the corresponding relative state of the operated vector.

A contraction semigroup is a strongly continuous semigroup of bounded linear operators satisfying $\Vert S(t)\Vert \le 1$ for every $t\ge 0$. The Hille-Yosida theorem characterizes its infinitesimal generator of a semigroup by dense domain, closedness, and the resolvent estimates $\Vert (\lambda -A{)}^{-n}\Vert \le {\lambda }^{-n}$ for real $\lambda >0$ and integers $n\ge 1$.

A bounded linear operator between normed spaces is contractive if its operator norm is at most one. This norm property is distinct from a strict contraction mapping, whose Lipschitz constant must be less than one.

A cyclic subspace generated by $x$ under a linear operator $T$ is the span of its successive iterates $x,Tx,T^{2}x,\dots$. In finite dimension it is already generated by the first $n$ iterates. The first linear dependence expresses the next iterate in previous ones and proves invariance without using the Cayley-Hamilton theorem.

For a linear operator $T$, a cyclic vector is a vector whose iterates span the entire vector space. In finite dimension $d$, the vectors $v,Tv,\dots ,T^{d-1}v$ then form a basis: the first linear dependence makes the preceding span invariant under $T$, so later iterates cannot enlarge it. In topological settings one often requires the iterates to have dense span instead; that closure convention must be stated.

An eigenspace of dimension greater than one. In degenerate perturbation theory, first-order energy corrections are the eigenvalues of the perturbing linear operator projected onto the whole unperturbed eigenspace.