= Hermitian form
{c}
{title2=$H(w,v)=\overline{H(v,w)}$}
A <Hermitian form> is a <sesquilinear form> on a <complex vector space> satisfying conjugate symmetry,
$$
H(w,v)=\overline{H(v,w)}.
$$
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 $A=A^*$: it is a <Hermitian matrix>. The coordinate rule remains $H(x,y)=x^T A\overline y$. Conjugate symmetry does not imply positivity.https://sites.math.washington.edu/~burke/crs/554/lectures/b-lecture4.pdf[Burke, Linear Analysis, Lecture 4, Hermitian symmetry]{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:
$$
B(u,v)=\frac14\bigl(q(u+v)-q(u-v)+i q(u+iv)-i q(u-iv)\bigr).
$$
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 $q(\alpha v)=|\alpha|^2q(v)$ 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 form>
$$
H_-(z,w)=z_1\overline{w_1}-z_2\overline{w_2},\qquad q_-(z)=|z_1|^2-|z_2|^2.
$$
Its 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 $4\leq0$. 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 $H_0(z,z)=|z_1|^2\geq0$, 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.
\Image[/sesquilinear-hermitian-forms-diagonal-slices.png]
{title=Analytical illustration: real slices of positive definite, indefinite, and positive semidefinite <Hermitian forms>}
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