Sesquilinear form

A complex sesquilinear form is linear in one argument and conjugate-linear in the other. Here the first argument is taken to be linear.

A Hermitian form is a complex sesquilinear form $H$ satisfying $H(w,v)=\overline{H(v,w)}$.

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 $H(x,y)=x^{T}A\overline{y}$.