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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.

For a bounded linear operator $T:H\to K$ between real or complex Hilbert spaces, its adjoint is the unique bounded linear operator $T^{\ast }:K\to H$ satisfyingThe inner product here is linear in its first argument. The adjoint operator transfers an output measurement back to the input: measuring $Tx$ against the vector $y$ gives exactly the same scalar as measuring $x$ against $T^{\ast }y$. It depends on both inner products.

In this real example,Both inner products equal $3$. The dashed lines collect all vectors giving that same measurement: the output line $u_1=3$ becomes the input line $2x_1+x_2=3$. The adjoint operator therefore moves the direction that describes the measurement. It generally differs from the inverse matrix: here $A^{-1}y=(1/2,0{)}^{\mathrm{T}}$.

For an explicitly complex example, use the standard inner product $⟨u,v⟩={\sum }_j{u}_j\overline{v_j}$ on ${\mathbb{C}}^2$. In the standard orthonormal basis, the adjoint operator is represented by the conjugate transpose:For $x=(1+i,2{)}^{\mathrm{T}}$ and $y=(1,i{)}^{\mathrm{T}}$, direct matrix multiplication givesand henceThe complex conjugate is essential to this agreement.

Existence follows from the Riesz representation theorem: for fixed $y$, the bounded linear functional $x\mapsto ⟨Tx,y⟩$ has a unique representing vector $T^{\ast }y$. The resulting adjoint operator has the same operator norm as $T$, andMIT's lecture 18 develops the representation argument and the conjugate transpose example.^[ref]

This transfer explains the normal equation for a linear inverse problem: a least-squares solution has residual $Tx-b$ orthogonal to every attainable direction $Th$, so $⟨h,T^{\ast }(Tx-b)⟩=0$ for every $h$, equivalently $T^{\ast }Tx=T^{\ast }b$. An output residual becomes an input condition through the same adjoint operator. All identities in this section concern bounded operators defined on the whole source Hilbert space; for a densely defined operator that is not bounded, use the adjoint of a densely defined operator and check its operator domain.

For a smooth ambient function near the unit sphere, the chain rule along the great circles ${\gamma }_i(t)=p\cos t+e_i\sin t$ gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-DF(p)$. Sum over an orthonormal basis of the tangent space and supplement it by the radial vector $p$. The geodesic trace formula for the Laplace-Beltrami operator then gives the displayed formula, evaluated at radius one. In particular a degree-$\ell$ harmonic polynomial restricts to a spherical harmonic with positive eigenvalue $\ell (\ell +d-1)$.

If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

An antiunitary operator between complex Hilbert spaces is a surjective antilinear map satisfying the displayed inner product identity. It preserves norms but conjugates scalars. In a chosen orthonormal basis it is $UK$, with $U$ a unitary operator and $K$ coefficientwise complex conjugation. Conversely any such $UK$ is antiunitary. Antilinearity by itself does not preserve norms.

A change of the chosen orthonormal basis of every anyon fusion space. In a multiplicity-free theory each basis change is an independent wave phase $u_c^{ab}$. Splitting vertices acquire $u_c^{ab}$ and adjoint fusion vertices acquire its inverse. A scalar diagram is invariant under these choices when its splitting and fusion vertex factors cancel. A closed diagram with independently specified unmatched vertex tensors need not be invariant.

If $g$ is gravitational acceleration and $a$ is the camera's acceleration, the locally felt upward vector is $U=a-g$. Choose the travel unit vector $X$, set $W=U-(U\cdot X)X$, and, when $W\ne 0$, take $Z=W/\Vert W\Vert$ and $Y=Z\times X$. These form an orthonormal basis with $U$ in the $XZ$ plane. If $U$ is parallel to $X$, the condition leaves the roll angle free even when $U\ne 0$; transport a previous transverse axis or choose a well-conditioned coordinate axis and project it into $X^{\perp }$.

Starting with $q_1=b/\Vert b{\Vert }_2$, the Arnoldi iteration constructs an orthonormal basis of successive Krylov subspaces. At step $j$, orthogonalize $A{q}_j$ against $q_1,\dots ,q_j$, and normalize the remainder as $q_{j+1}$ if it is nonzero. The coefficients form an upper Hessenberg matrix ${\tilde{H}}_k$ and satisfy $A{Q}_k=Q_{k+1}{\tilde{H}}_k$. A zero remainder means the current Krylov subspace is invariant under $A$.

For a fixed nonzero real vector $b$ in three dimensions, a real linear map commuting with every proper rotation about that axis has the displayed form, where $[b{]}_{\times }x=b\times x$ is the cross-product matrix. On the axis it acts by $a+c\Vert b{\Vert }^2$. On the perpendicular plane its matrix in an oriented orthonormal basis is $\left(\begin{array}{cc}a& -d\Vert b\Vert \\ d\Vert b\Vert & a\end{array}\right)$. Its eigenvalues are therefore $a+c\Vert b{\Vert }^2$ and $a\pm id\Vert b\Vert$, andTo see the converse, commuting with all rotations preserves their common fixed axis and its perpendicular plane; there can be no nonzero coupling vector fixed by every planar rotation. A planar matrix commuting with every rotation is a scalar identity plus a scalar right-angle rotation. Thus the block description is exhaustive. If $d\ne 0$, only the axial eigenvalue is real; if $d=0$, the perpendicular plane also consists of real eigenvectors.

If a rigid body's mass density is invariant under every rotation about an axis through its centre of mass, its inertia tensor in any orthonormal basis with third vector on that axis has the displayed form. In particular all transverse directions are principal axes with equal principal moments of inertia.

For a proof, let that axis be $z$. In the defining integrals for the inertia tensor, a half-turn $(x,y,z)\mapsto (-x,-y,z)$ makes the $xz$ and $yz$ products change sign, so their integrals vanish. A quarter-turn $(x,y,z)\mapsto (-y,x,z)$ makes the $xy$ integral vanish and equates the integrals of $x^2$ and $y^2$. Therefore all off-diagonal components vanish and $I_{11}=I_{22}$. The values are $I_{\perp }=\int \rho [(x^2+y^2)/2+z^2]d^{3}r$ and $I_{\parallel }=\int \rho (x^2+y^2)d^{3}r$. No reflection symmetry in $z$ is needed.

The Bargmann-Fock space is a Hilbert space of entire holomorphic functions with Gaussian-weighted area inner product. If $f(z)=\sum a_n{z}^n$, angular integration and the Gaussian integral give $\Vert f{\Vert }^2=\sum n!|a_n{|}^2$. Conversely any such square-summable coefficient sequence defines an entire function, since $\sum |a_n{z}^n|\le (\sum n!|a_n{|}^2{)}^{1/2}{e}^{|z{|}^2/2}$. This proves completeness and shows that $z^n/\sqrt{n!}$ is an orthonormal basis. The same bound shows boundedness of evaluation and gives reproducing kernel $e^{z\bar{w}}$.

A rank-one projective measurement associated with an orthonormal basis $\{|b_j⟩{\}}_j$. For a density matrix $\rho$, outcome $j$ has probability $⟨b_j|\rho |b_j⟩$ and leaves the measured system in the pure state $|b_j⟩$ whenever that probability is nonzero. It can be implemented by mapping this orthonormal basis to the computational basis with a unitary transformation and performing a computational-basis measurement.^[ref]

A basis is a linearly independent spanning set of vectors in a vector space. Every vector is a unique finite linear combination of basis vectors. Independence supplies uniqueness, and spanning supplies existence. An ordered basis turns the coefficients into a coordinate vector.

For $\mathcal{B}=(b_1,b_2)=((2,1),(1,2))$, the matrix with these vectors as columns isHence the coordinate vector of $v=(5,4)$ is $[v{]}_{\mathcal{B}}=B^{-1}v=(2,1)$, and $B[v{]}_{\mathcal{B}}=v$ recovers the usual coordinates. The determinant $3$ is nonzero, so this recovery works for every vector in ${\mathbb{R}}^2$. A basis need not have perpendicular or unit-length vectors; an orthonormal basis adds those geometric conveniences when an inner product is available.

The point of changing basis is to simplify the operation being studied. Diagonalization uses an eigenbasis to turn a coupled linear map into independent scalings. In a space of polynomials, the basis $1,x,x^2$ makes differentiation easy to encode, while orthogonal polynomials can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.^[ref]

Relative to a fixed orthonormal basis, the displayed linear map is an isometry because its basis images are orthonormal. Its adjoint obeys $C^{\ast }|k⟩|\ell ⟩={\delta }_{k\ell }|k⟩$. If vectors are described by coefficient functions, this adjoint maps $|f⟩|g⟩$ to $|fg⟩$, where the product is pointwise. Although $C$ copies every basis quantum state, it does not copy arbitrary superpositions. For example it sends $(|0⟩+|1⟩)/\sqrt{2}$ to $(|00⟩+|11⟩)/\sqrt{2}$, whereas two copies would also contain the cross terms $|01⟩$ and $|10⟩$.

For any real separable Hilbert space $H$, any orthonormal basis $(e_j)$ and independent standard normal random variables $(Z_j)$, the displayed series converges almost surely for each fixed $h\in H$ and in L2 space. Indeed, the Parseval identity for a Hilbertian basis gives ${\sum }_j⟨h,e_j{⟩}^2=\Vert h{\Vert }^2$, so the partial sums form an L2 space-bounded martingale. The martingale convergence theorem supplies both limits. Every finite linear combination has the normal distribution with variance $\Vert {\sum }_j{u}_j{h}_j{\Vert }^2$. Thus its finite-dimensional characteristic function is $\exp (-\Vert {\sum }_j{u}_j{h}_j{\Vert }^2/2)$, independent of the chosen orthonormal basis. This proves equality of the finite-dimensional distributions of the resulting isonormal Gaussian processes; pathwise equality needs a coupling.

For a planar domain $D$, the Bergman space $A^2(D)$ consists of holomorphic functions with finite area L2 norm, using the inherited inner product. It is a closed subspace of $L^2(D)$. Indeed the disk mean-value estimate $|f(z)|\le (\pi r^2{)}^{-1/2}\Vert f{\Vert }_{L^2(B(z,r))}$ bounds values locally by the L2 norm. An $L^2$-Cauchy sequence of holomorphic functions consequently converges uniformly on compact subsets to a holomorphic function, agreeing almost everywhere with its $L^2$ limit.

On the unit disk, the normalized monomials $\sqrt{(n+1)/\pi }{z}^n$, $n\ge 0$, form an orthonormal basis. Angular integration makes them orthogonal and gives $\Vert z^n{\Vert }_2^2=\pi /(n+1)$. Applying the same integration to a Taylor expansion gives $\Vert {\sum }_n{a}_n{z}^n{\Vert }_2^2=\pi {\sum }_n|a_n{|}^2/(n+1)$, proving completeness of this basis.

The best N-term approximation in an orthonormal basis retains $N$ coefficients of greatest absolute value. It minimizes the squared Hilbert space error over all index sets of size at most $N$. The selection makes the approximation nonlinear in the data, even though coefficient extraction is linear.

The Born rule connects a quantum state to measurement probabilities. Let $\{|j⟩\}$ be a complete orthonormal basis, finite or countably infinite, for a Hilbert space, and letAn ideal measurement distinguishing these basis vectors returns label $j$ with probabilityThis assignment is a physical measurement rule. Its compatibility with total probability one follows from orthonormality and normalization:The probabilities are the squared moduli, not the amplitudes themselves or the squares $a_j^2$. ^[ref].

If the basis is an eigenbasis of an observable, the reported numerical value is its eigenvalue. Several vectors can share one eigenvalue; the probability of that value is then the sum over the corresponding basis labels. More generally, a projective measurement with orthogonal projections $P_r$ assignsfor a pure state and a general density matrix $\rho$, respectively. This also covers mixed states and degenerate eigenspaces. The Born rule determines outcome statistics; the ideal conditional state change is specified separately by the Lüders rule. See projective measurement for that formulation and its assumptions. ^[ref].

For a worked example, take the normalized qubit quantum stateThe squared moduli add to $3/4+1/4=1$. A computational-basis measurement, equivalently a measurement of the $Z$ Pauli matrix, hasThese probabilities do not depend on the relative quantum phase $\varphi$. Now choose the Hadamard basiswhose vectors are eigenstates of the $X$ Pauli matrix, also used by the Pauli X gate. The new probability amplitudes and probabilities areBoth probabilities are nonnegative and their sum is one. The relative quantum phase becomes visible because the two probability amplitudes add before their squared modulus is taken:

A Hadamard gate followed by a computational-basis measurement realizes these $X$ statistics, since $H|\pm ⟩=|0⟩,|1⟩$, respectively. Compare the two bases using fresh, identically prepared copies: a preceding measurement may change the state. The quantum superposition section shows why replacing the preparation by the mixed state $\mathrm{diag}(3/4,1/4)$ keeps the $Z$ statistics but gives $p_{+}=p_{-}=1/2$ for every $\varphi$.

The usefulness of the Born rule is this conversion from linear algebra to experimental predictions: calculate overlaps in the basis actually measured, then take squared moduli. Its appealing geometry is that changing basis redistributes probability while preserving total squared norm. The same rule relates an energy eigenstate expansion to an energy measurement and a position wavefunction to a probability density.

Born's 1926 paper on quantum scattering interpreted the outgoing-wave coefficients in terms of the probabilities of collision outcomes; a corrective footnote specified proportionality to the square of the amplitude. This is a historical scattering formulation, rather than the general density matrix formula above. ^[ref]; ^[ref].

For a complete orthonormal basis of $L^2({\mathbb{R}}_{+})$, the Parseval identity for a Hilbertian basis identifies the sum of the feature products with $⟨1_{[0,s]},1_{[0,t]}⟩=\min (s,t)$. The Cauchy-Schwarz inequality gives absolute convergence.

For a centered Gaussian measure on a real Hilbert space with injective covariance operator of a Gaussian measure $\Sigma$, its Cameron-Martin space is $E_{\mu }=\mathrm{Ran}{\Sigma }^{1/2}$, with inner product $⟨h,g{⟩}_{E_{\mu }}=⟨{\Sigma }^{-1/2}h,{\Sigma }^{-1/2}g⟩$. In an orthonormal basis satisfying $\Sigma e_j={\lambda }_j{e}_j$, it consists of exactly the vectors $h$ for which ${\sum }_j{h}_j^2/{\lambda }_j<\infty$. It is complete in this stronger norm even if its range is not closed in the ambient norm. Its importance is that it describes precisely the translations that preserve the null sets of the Gaussian measure, as in the Cameron-Martin theorem for a Gaussian measure. For generalized Gaussian white noise over $L^2$, the corresponding Cameron-Martin space is L2 space itself.