Hermitian form ($H(w,v)=\overline{H(v,w)}$)

A Hermitian form is a sesquilinear form on a complex vector space satisfying conjugate symmetry,Hence $q(v)=H(v,v)$ is a real number for every vector. In a basis, the matrix of a Hermitian form has $A_{ij}=H(e_i,e_j)$ and obeys $A_{ji}=\overline{A_{ij}}$, equivalently $A=A^{\ast }$: it is a Hermitian matrix. The coordinate rule remains $H(x,y)=x^{T}A\overline{y}$. Conjugate symmetry does not imply positivity.^[ref]

The diagonal contains more information than might be expected. For any complex sesquilinear form $B$, put $q(v)=B(v,v)$. Expanding four expressions gives the complex polarization identity in the linear-first convention:This is an algebraic identity, with no positivity assumption. If every $q(v)$ is real, interchanging $u,v$ preserves the first difference and negates the second: use $q(\alpha v)=|\alpha {|}^{2}q(v)$ for $\alpha =-1,i$. The identity then gives $B(v,u)=\overline{B(u,v)}$. Thus a complex sesquilinear form is a Hermitian form if and only if all its diagonal values are real. Testing only the basis vectors is insufficient: the off-diagonal example in sesquilinear form has zero diagonal entries in its matrix but a nonreal value at $(1,i)$.

Consider the indefinite Hermitian formIts values at $e_1,e_2$ are $1,-1$. Moreover $u=(1,1)$ and $v=(1,-1)$ are nonzero isotropic vectors: $q_{-}(u)=q_{-}(v)=0$, yet $H_{-}(u,v)=2$. These vectors do not belong to the radical of a Hermitian form, and the usual Cauchy-Schwarz inequality would incorrectly require $4\le 0$. An indefinite Hermitian form records signed geometry rather than defining a norm by $\sqrt{H(v,v)}$.

There is a separate boundary case: $H_0(z,w)=z_1\overline{w_1}$ is a positive semidefinite Hermitian form, since $H_0(z,z)=|z_1{|}^2\ge 0$, but $H_0(e_2,e_2)=0$. Its radical of a Hermitian form is the entire second coordinate line. Positive definiteness is exactly the additional condition in a Hermitian inner product. The three examples below show why these distinctions matter: on the real slice $z=(x,y)\in {\mathbb{R}}^2\subset {\mathbb{C}}^2$, $q_{+}(z)=|z_1{|}^2+2|z_2{|}^2$ has an ellipse as its unit level, $q_{-}$ has hyperbolas, and $q_0$ has two parallel lines. These are slices of the complex examples, not drawings of their full four-dimensional domains.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

Each basis $e_1,\dots ,e_n$ of a finite-dimensional complex vector space determines a unique complex inner product making that basis orthonormal. In coordinates it is $⟨{\sum }_i{u}_i{e}_i,{\sum }_i{v}_i{e}_i{⟩}_e={\sum }_i{u}_i\overline{v_i}$, with the first argument linear. The formula is sesquilinear, has conjugate symmetry, and satisfies $⟨v,v{⟩}_e={\sum }_i|v_i{|}^2$, which is positive unless $v=0$. Expanding any inner product in the given basis proves uniqueness. Its codomain is the complex numbers; only its diagonal values are necessarily real. The resulting identification with the dual space is conjugate-linear, not complex-linear.

Exchanging the arguments of a Hermitian inner product takes its complex conjugate. This property, together with sesquilinearity and positive definiteness, distinguishes a complex inner product from a general bilinear pairing.

Over the finite field $K={\mathbb{F}}_{q^2}$, a sesquilinear form relative to $a\mapsto a^q$ is linear in its first argument and conjugate-linear in its second. It is Hermitian when $B(w,v)=\overline{B(v,w)}$. Its diagonal values belong to ${\mathbb{F}}_q$; it is nondegenerate when $B(v,w)=0$ for every $w$ forces $v=0$. This definition applies in characteristic two as well.

In a two-dimensional nondegenerate Hermitian form over a quadratic finite field, choose an orthonormal basis $v,w$ and $a$ with $a\bar{a}=-1$. Then $e=v+aw$ is an isotropic vector. Choose $g$ with $B(e,g)=1$, and choose $c$ with $c+\bar{c}=B(g,g)$. The vector $f=g-ce$ satisfies $B(e,f)=1$ and $B(f,f)=0$. Thus $e,f$ give the displayed matrix, including in characteristic two. Iterating on nondegenerate orthogonal complement for a sesquilinear form leaves only a unit-norm line in odd dimension.

A nondegenerate Hermitian form over a quadratic finite field has an orthonormal basis. If all diagonal values vanished, evaluating $B(v+aw,v+aw)$ would give $\mathrm{Tr}(\bar{a}B(v,w))=0$ for every $a$, contradicting quadratic finite-field norm and trace surjectivity and nondegeneracy. Thus some $v$ has nonzero norm; rescaling using the field norm makes it one. Its orthogonal complement for a sesquilinear form is nondegenerate, so induction completes the basis.

Integrate a polynomial times the conjugate of another over the product of unit circles, with normalized product angular measure. Fourier orthogonality makes the monomials $z_1^{{\alpha }_1}\cdots z_d^{{\alpha }_d}$ an orthonormal basis of the polynomial vector space. The Hermitian form is positive definite because the squared norm is the sum of squared absolute values of the finitely many coefficients. No completion or infinite series is needed to obtain this algebraic basis.

The radical of a Hermitian form is the vector subspace orthogonal to the entire space. The form descends to the quotient vector space by its radical: adding a radical vector to either argument does not change the value. For a positive semidefinite Hermitian form, the Cauchy-Schwarz inequality identifies this radical with its zero-norm vectors, so the quotient form is positive definite. This conclusion does not hold for a general indefinite Hermitian form: a zero-norm vector need not belong to its radical.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

An indefinite Hermitian form takes both positive and negative values on $⟨v,v⟩$ for nonzero vectors. For example, $-|v_0{|}^2+|v_1{|}^2+|v_2{|}^2+|v_3{|}^2$ is indefinite. It is not a positive inner product defining a Hilbert space, and a nonzero vector can have zero norm without being orthogonal to all vectors. A positive semidefinite restricted form becomes a positive inner product only after quotienting its radical of a Hermitian form.

For a basis $(v_i)$, the matrix $A$ of a Hermitian form has entries $A_{ij}=H(v_i,v_j)$. With the convention that $H$ is linear in its first argument, coordinate columns $x,y$ satisfy $H(x,y)=x^{T}A\overline{y}$.