Inner product: Difference between revisions
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It must satisfies four properties: | It must satisfies four properties: | ||
* Linearity in | * Linearity in the 1st slot: $\langle u + v, w \rangle = \langle u, w \rangle + \langle v, w \rangle$ and $\langle au, v \rangle = a \langle u, v \rangle$ for all $u, v, w \in V; a \in \mathbb{F}$. | ||
* Positivity: $\langle v, v \rangle \geq 0$ for all $v \in V$. | * Positivity: $\langle v, v \rangle \geq 0$ for all $v \in V$. | ||
* Positive definiteness: $\langle v, v \rangle = 0$ if and only if $v = 0$. | * Positive definiteness: $\langle v, v \rangle = 0$ if and only if $v = 0$. | ||
Revision as of 14:20, 29 September 2026
An inner product on $V$ is a map
$$ \langle \cdot, \cdot \rangle : V \times V \to \mathbb{F}, \quad (u, v) \mapsto \langle u, v \rangle $$
It must satisfies four properties:
- Linearity in the 1st slot: $\langle u + v, w \rangle = \langle u, w \rangle + \langle v, w \rangle$ and $\langle au, v \rangle = a \langle u, v \rangle$ for all $u, v, w \in V; a \in \mathbb{F}$.
- Positivity: $\langle v, v \rangle \geq 0$ for all $v \in V$.
- Positive definiteness: $\langle v, v \rangle = 0$ if and only if $v = 0$.
- Conjugate symmetry: $\langle u, v \rangle = \overline{\langle v, u \rangle}$ for all $u, v \in V$.
Recall $$\forall x\in\mathbb{R},\; x=\overline{x}$$. Hence, for real vector spaces, conjugate symmetry of an inner product becomes actual symmetry.