Inner product

Definition

Given a vector space, $(V,F)$ (where $F$ is either $\mathbb{R}$ or $\mathbb{C}$ ), an inner product^[1]^[2]^[3] is a map:

$\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{R}$ (or sometimes $\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{C}$ )

Such that:

$\langle x,y\rangle = \overline{\langle y, x\rangle}$ (where the bar denotes Complex conjugate)
- Or just if the inner product is into $\mathbb{R}$
$\langle\lambda x+\mu y,z\rangle = \lambda\langle y,z\rangle + \mu\langle x,z\rangle$ ( linearity in first argument )
This may be alternatively stated as:
- $\langle\lambda x,y\rangle=\lambda\langle x,y\rangle$ and $\langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle$
$\langle x,x\rangle \ge 0$ but specifically:
- $\langle x,x\rangle=0\iff x=0$

Notice that $\langle\cdot,\cdot\rangle$ is also linear (ish) in its second argument as:

$\langle x,\lambda y+\mu z\rangle = \overline{\langle \lambda y+\mu z, x\rangle}$ $=\overline{\lambda\langle y,x\rangle + \mu\langle z,x\rangle}$ $=\bar{\lambda}\overline{\langle y,x\rangle}+\bar{\mu}\overline{\langle z,x\rangle}$ $=\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle$

From this we may conclude the following:

This leads to the most general form:

$\langle au+bv,cx+dy\rangle=a\langle u,cx+dy\rangle+b\langle v,cx+dy\rangle$ $=a\overline{\langle cx+dy,u\rangle}+b\overline{\langle cx+dy,v\rangle}$ $=a(\overline{c\langle x,u\rangle} + \overline{d\langle y,u\rangle})+b(\overline{c\langle x,v\rangle}+\overline{d\langle y,v\rangle})$ $=a\overline{c}\langle u,x\rangle+a\overline{d}\langle u,y\rangle+b\overline{c}\langle v,x\rangle+b\overline{d}\langle v,y\rangle$

Proof of claim: $\langle x,\alpha y+\beta z\rangle=\overline{\alpha}\langle x, y\rangle +\overline{\beta}\langle x,z\rangle$