= Hermitian inner product
{c}
{title2=$H(v,v)>0\quad(v\ne0)$}
= Complex inner product
{synonym}
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 does
$$
\|v\|=\sqrt{H(v,v)}
$$
define 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 $\sum_j c_jz_j\overline{w_j}$, with squared <norm> $\sum_j c_j|z_j|^2$.https://linear.axler.net/InnerProduct.pdf[Sheldon Axler, Linear Algebra Done Right, third edition, §6.A, definitions 6.3 and 6.9 and examples 6.4]{ref}
A <Hermitian inner product> need not have a real <matrix>. For instance, with the linear-first coordinate convention,
$$
A=\begin{pmatrix}2&i\\-i&2\end{pmatrix},\qquad H(z,w)=z^T A\overline w,\qquad
H(z,z)=|z_1-i z_2|^2+|z_1|^2+|z_2|^2.
$$
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 $z^*Az$ 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> states
$$
|H(v,u)|^2\leq H(v,v)H(u,u).
$$
Equality 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> gives
$$
H(v,v)=\frac{|H(v,u)|^2}{H(u,u)}+H(r,r).
$$
Nonnegativity 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.https://linear.axler.net/InnerProduct.pdf[Axler, Linear Algebra Done Right, third edition, orthogonal decomposition 6.14 and Cauchy-Schwarz inequality 6.15]{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.
\Image[/sesquilinear-hermitian-forms-projection-phase.png]
{title=Analytical illustration: weighted <orthogonal projection> on a real slice and opposite changes of <complex argument> of its projection coefficient}
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>.
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