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The exponential series converges absolutely in a unital Banach algebra. Multiplication of absolutely convergent series gives $e^x{e}^{-x}=1$, and gives $e^{x+y}=e^x{e}^y$ when $x,y$ commute. Termwise differentiation gives $e^{tx}=1+tx+O(t^2)$ near zero. In a C-star algebra, if $h=h^{\ast }$ and $t$ is real, taking adjoints termwise gives $(e^{ith}{)}^{\ast }=e^{-ith}$, so this exponential is a Unitary element of a C-star algebra and has norm one. These facts let a norm bound on a functional recover its real values on Hermitian elements.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

For a Normal element of a C-star algebra $a$ in a unital C-star algebra, the Commutative Gelfand--Naimark theorem identifies $C(\sigma (a))$ isometrically with the closed unital algebra generated by $a$ and $a^{\ast }$. The coordinate function maps to $a$, defining $f(a)$ for every continuous $f$ on the spectrum of an element.

A C-star algebra is a complex Banach algebra with an algebra involution $a\mapsto a^{\ast }$ satisfying the C-star identityThe algebra involution is conjugate-linear, reverses products and obeys $(a^{\ast }{)}^{\ast }=a$. An identity need not be present.^[ref]

Examples include complex matrices with the operator norm and conjugate transpose, bounded linear operators on a Hilbert space with the operator adjoint, and $C_0(X)$ on a Hausdorff space that is a locally compact space with pointwise complex conjugation and the supremum norm. The identity ties the norm to the algebraic product and involution: it implies spectral radius norm equality for normal elements and makes an injective C-star homomorphism an isometry. This rigidity permits continuous functional calculus for normal C-star elements.

A C-star homomorphism is a complex linear multiplicative map between C-star algebras that preserves the involution. Such maps are contractive; an injective one is an isometry.

A C-star isomorphism is a bijective C-star homomorphism: a complex-linear multiplicative map preserving the involution. Between unital C-star algebras it automatically preserves the unit. It preserves the Banach algebra spectrum, since it preserves invertibility in both directions. Applying the C-star identity and the spectral norm formula to $a^{\ast }a$ gives $\Vert \Phi (a){\Vert }^2=\Vert \Phi (a^{\ast }a)\Vert =\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$, so it is automatically isometric.

A C-star subalgebra is a norm-closed subalgebra closed under the involution, with the inherited C-star identity. A unital inclusion in this context uses the same identity as the ambient C-star algebra. The spectral permanence for C-star algebras says the inclusion preserves the spectrum of an element.

The left multiplication representation of a C-star algebra is isometric: test $L_a$ on $a^{\ast }/\Vert a\Vert$ to get $\Vert L_a\Vert =\Vert a\Vert$. For a nonunital algebra, its closed image plus the identity operator is the algebraic unitization. For a formal $z=a+\lambda 1$ and $b\in A$, $\Vert zb{\Vert }^2=\Vert b^{\ast }{z}^{\ast }zb\Vert \le \Vert b\Vert \Vert L_{z^{\ast }z}b\Vert$. Thus $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$; applying the same estimate to $z^{\ast }$ proves the algebra involution is isometric and the C-star identity holds. This gives a unitization without first assuming an operator representation on a Hilbert space.

A representation of a C-star algebra on a Hilbert space is a star-homomorphism $\pi :A\to \mathcal{B}(H)$. It is cyclic if some vector $\xi$ has dense orbit span as displayed; the orbit $\pi (A)\xi$ is itself a linear subspace. For a unital representation, a unit cyclic vector defines a state on a C-star algebra by $a\mapsto ⟨\pi (a)\xi ,\xi ⟩$. Conversely, the Gelfand–Naimark–Segal construction realizes every state this way. Cyclicity means density under the action of the whole algebra, not necessarily under powers of a single chosen element.

For a nonzero unital C-star algebra, a positive functional on a C-star algebra obeys $\Vert x\Vert =x(1_A)$, and conversely this equality for a bounded functional implies positivity. Therefore its state on a C-star algebra functionals are exactly the maximizers of the real linear evaluation $x\mapsto \mathrm{Re}x(1_A)$ on the closed dual unit ball. If a convex combination reaches that maximum, each constituent must reach it: each real part is at most one, and their weighted mean is one. The modulus bound then forces evaluation to equal one, so each constituent is again a state. Hence the state space is an exposed face and a face of a convex set.

For density operators $\rho ,\sigma$ on the same finite-dimensional Hilbert space, quantum fidelity measures how close their quantum states are. This article uses the unsquared convention:Each square root is the positive square root of an operator, and $\Vert \cdot {\Vert }_1$ is the trace norm. Quantum fidelity is symmetric in its inputs; it is one exactly when $\rho =\sigma$, and zero exactly when $\rho \sigma =0$, meaning that the support of a positive operator for one state is orthogonal to that for the other. Larger quantum fidelity means greater similarity. The separate squared quantum fidelity is $F_{\mathrm{sq}}=F^2$. ^[ref]

For two normalized pure states, quantum fidelity is the magnitude of their inner product, $F=|⟨\psi |\phi ⟩|$. Thus $|0⟩$ compared with itself has $F=1$, while $|0⟩$ and $|1⟩$ have $F=0$. For $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$, comparison with $|0⟩$ gives $F=1/\sqrt{2}$, $F_{\mathrm{sq}}=1/2$, and trace distance $D=1/\sqrt{2}$ by the pure-state trace distance and fidelity identity.

Why compare density operators rather than their entries? A change of coordinates should not make the same physical preparations look more or less alike. Unitary invariance of trace distance and fidelity makes quantum fidelity independent of a common basis choice. Multiplying a pure state vector by a global phase leaves its density operator unchanged, so its quantum fidelity with the original state is one. These features make quantum fidelity useful for checking whether quantum channels, state preparations, and Schumacher compression preserve a target quantum state.

The connection to Uhlmann's theorem supplies another beautiful interpretation: quantum fidelity is the largest absolute inner product between pure vectors obtained from a purification of a density operator for each state, using a sufficiently large common reference space. Uhlmann's 1976 paper extended transition probability to algebraic states, including those of a C-star algebra; for density operators its transition probability corresponds to our squared quantum fidelity. The measurement characterization of quantum fidelity gives a complementary description directly in terms of observable outcome statistics. ^[ref]

A finite family of pairwise commuting normal operators on a finite-dimensional complex inner product space, commuting also with each other's adjoints, generates a commutative unital C-star algebra. Its finite maximal ideal space identifies the algebra with functions on finitely many points. The characteristic functions of those points pull back under the Gelfand transform to mutually orthogonal self-adjoint linear projections summing to the identity. Their ranges are the simultaneous eigenspaces, giving simultaneous diagonalization without choosing separate incompatible eigenbases.

Let $F$ be a nonempty norm-closed face of a convex set $[0,1_A]$ in a unital C-star algebra, with $0\notin F$. Every $a\in F$ has norm one: otherwise $a=\Vert a\Vert (a/\Vert a\Vert )+(1-\Vert a\Vert )0$ is a nontrivial convex decomposition forcing zero into the face. If $g:[0,1]\to [0,1]$ is continuous and nondecreasing with $g(0)=0,g(1)=1$, then $g(a)\in F$.

To prove this, for $n\ge 2$ replace $g$ near the endpoints by a function $g_n$ which is zero on $[0,1/(2n)]$, one on $[1-1/(2n),1]$, agrees with $g$ on $[1/n,1-1/n]$, and interpolates linearly in between. Monotonicity and endpoint continuity give $g_n\to g$ uniformly. With $\delta =1/(2n)$, the function $h_n(t)=(t-\delta g_n(t))/(1-\delta )$ takes values in $[0,1]$: check the two constant regions and use $\delta \le t\le 1-\delta$ elsewhere. The continuous functional calculus gives $a=\delta g_n(a)+(1-\delta )h_n(a)$ with both summands positive contractions. The face property gives $g_n(a)\in F$, and closedness gives $g(a)\in F$.

For a commutative complex unital Banach algebra, its Gelfand transform is a continuous unital algebra homomorphism into the continuous functions on its compact character space. The values of the transform of $a$ are exactly ${\sigma }_A(a)$, so $\Vert \hat{a}{\Vert }_{\infty }=r(a)$. Its kernel is the intersection of the kernels of the algebra characters. For a commutative C-star algebra, spectral radius norm equality for normal elements and the Stone-Weierstrass theorem upgrade this to an isometric C-star homomorphism onto $C(\Delta (A))$.

A Hermitian Banach algebra is a Banach algebra with an algebra involution for which every Hermitian element of a star algebra has real spectrum of an element. A closed unital star-subalgebra has spectral permanence for Hermitian Banach algebras. The assumption is weaker than being a C-star algebra and does not require the C-star norm identity.

An element $x$ of a C-star algebra is Hermitian when $x=x^{\ast }$. Its spectrum is real, its norm equals its spectral radius, and it is the difference of two positive elements.

For a strongly continuous unitary representation, integration against $f\in L^1(G)$ defines an operator satisfying $\Vert \pi (f)\Vert \le \Vert f{\Vert }_1$. Fubini's theorem gives $\pi (f\ast g)=\pi (f)\pi (g)$, and the group-algebra involution gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. On an Abelian group, the operator-norm closure is a commutative C-star algebra. Its algebra characters produce continuous unitary group characters, allowing the Commutative Gelfand--Naimark theorem and Riesz-Markov-Kakutani representation theorem to turn a positive vector functional into a finite spectral measure.

An element $x$ of a C-star algebra is normal when $x^{\ast }x=x{x}^{\ast }$.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

The continuous functional calculus for a Hermitian element of a C-star algebra identifies its generated unital C-star algebra with continuous functions on its real spectrum, preserves the involution and norm, and sends functions to the corresponding spectral values by spectral mapping. Consequently an element is a square of a Hermitian element of a C-star algebra exactly when it is a Hermitian element of a C-star algebra with nonnegative spectrum: the reverse direction uses the continuous square root on that spectrum.

Nonnegative scalar multiples preserve this property directly. For positive $a,b$ put $C=\Vert a\Vert +\Vert b\Vert$. Functional calculus gives $\Vert \Vert a\Vert 1-a\Vert \le \Vert a\Vert$ and similarly for $b$, so $\Vert C1-(a+b)\Vert \le C$. The sum is a Hermitian element of a C-star algebra, and spectral mapping now places its spectrum in $[0,2C]$. It therefore has a positive square root in a C-star algebra. If $C=0$ both elements are zero. Thus $A_{+}$ is a cone. It is pointed because an element positive with its negative has spectrum contained in $\{0\}$ and hence is zero by the self-adjoint norm formula. It is closed as well: for a converging sequence of Positive elements of a C-star algebra use a common norm bound $C$ and pass to the limit in $\Vert C1-a_n\Vert \le C$.

For the two-by-two matrix the exact criterion isA positive matrix is Hermitian with nonnegative eigenvalues; evaluating its quadratic form at the coordinate vectors gives the diagonal conditions, and its determinant, the eigenvalue product, gives the final inequality. Conversely if $a_{11}>0$, completing the square writes the quadratic form as $a_{11}|z_1+(a_{12}/a_{11})z_2{|}^2+(a_{22}-|a_{12}{|}^2/a_{11})|z_2{|}^2$, which is nonnegative. If $a_{11}=0$, the determinant inequality forces $a_{12}=0$ and the conclusion is immediate. The spectral theorem then gives nonnegative eigenvalues and their Hermitian square root, proving positivity in the question's sense.