mathematics-physics-wiki/docs/en/mathematics/linear-algebra/inner-product-spaces.md

# Inner product spaces 

## Definition

An introduction of length in a vector space may be formulated in terms of an inner product space. 

> *Definition 1*: an **inner product** on $V$ is an operation on $V$ that assigns, to each pair of vectors $\mathbf{x},\mathbf{y} \in V$, a real number $\langle \mathbf{x},\mathbf{y}\rangle$ satisfying the following conditions
>
> 1. $\langle \mathbf{x},\mathbf{x}\rangle > 0, \text{ for } \mathbf{x} \in V\backslash\{\mathbf{0}\} \text{ and }  \langle \mathbf{x},\mathbf{x}\rangle = 0, \; \text{for } \mathbf{x} = \mathbf{0}$,
> 2. $\langle \mathbf{x},\mathbf{y}\rangle = \overline{\langle \mathbf{y},\mathbf{x}\rangle}, \; \forall \mathbf{x}, \mathbf{y} \in V$, 
> 3. $\langle a \mathbf{x} + b \mathbf{y}, \mathbf{z}\rangle = a \langle \mathbf{x},\mathbf{z}\rangle + b \langle \mathbf{y},\mathbf{z}\rangle, \; \forall \mathbf{x}, \mathbf{y}, \mathbf{z} \in V \text{ and } a,b \in \mathbb{K}$.

A vector space $V$ with an inner product is called an **inner product space**. 

### Euclidean inner product spaces

The standard inner product on the Euclidean vector spaces $V = \mathbb{R}^n$ with $n \in \mathbb{N}$ is given by the scalar product defined by

$$
    \langle \mathbf{x},\mathbf{y}\rangle = \mathbf{x}^T \mathbf{y}, 
$$

for all $\mathbf{x},\mathbf{y} \in V$. 

??? note "*Proof*:"

    Will be added later.

This can be extended to matrices $V = \mathbb{R}^{m \times n}$ with $m,n \in \mathbb{N}$ for which an inner product may be given by

$$
    \langle A, B\rangle = \sum_{i=1}^m \sum_{j=1}^n a_{ij} b_{ij}, 
$$

for all $A, B \in V$. 

??? note "*Proof*:"

    Will be added later.

### Function inner product spaces

Let $V$ be a function space with a domain $X$. An inner product on $V$ may be defined by

$$
    \langle f, g\rangle = \int_X \bar f(x) g(x) dx 
$$

for all $f,g \in V$. 

??? note "*Proof*:"

    Will be added later.

### Polynomial inner product spaces

Let $V$ be a polynomial space of degree $n \in \mathbb{N}$ with the set of numbers $\{x_i\}_{i=1}^n \subset \mathbb{K}^n$. An inner product on $V$ may be defined by

$$
    \langle p, q \rangle = \sum_{i=1}^n \bar p(x_i) q(x_i),
$$

for all $p,q \in V$. 

??? note "*Proof*:"

    Will be added later.

## Properties of inner product spaces

> *Definition 2*: let $V$ be an inner product space, the Euclidean length $\|\mathbf{v}\|$ of a vector $\mathbf{v}$ is defined as
>
> $$
>   \|\mathbf{v}\| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle},
> $$
>
> for all $\mathbf{v} \in V$. 

Which is consistent with Euclidean geometry. According to definition 1 the distance between two vectors $\mathbf{v}, \mathbf{w} \in V$ is $\|\mathbf{v} - \mathbf{w}\|$.

> *Definition 3*: let $V$ be an inner product space, the vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if
>
> $$
>   \langle \mathbf{u}, \mathbf{v} \rangle = 0,
> $$
>
> for all $\mathbf{u}, \mathbf{v} \in V$. 

A pair of orthogonal vectors will satisfy the theorem of Pythagoras.

> *Theorem 1*: let $V$ be an inner product space and $\mathbf{u}$ and $\mathbf{v}$ are orthogonal then
>
> $$
>   \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2,
> $$
>
> for all $\mathbf{u}, \mathbf{v} \in V$. 

??? note "*Proof*:"

    let $V$ be an inner product space and let $\mathbf{u}, \mathbf{v} \in V$ be orthogonal, then

    $$
    \begin{align*}
        \|\mathbf{u} + \mathbf{v}\|^2 &= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v}\rangle, \\
                                    &= \langle \mathbf{u}, \mathbf{u} \rangle + 2 \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle, \\
                                    &= \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2. 
    \end{align*}
    $$

Interpreted in $\mathbb{R}^2$ this is just the familiar Pythagorean theorem.

> *Definition 4*: let $V$ be an inner product space then the **scalar projection** $a$ of $\mathbf{u}$ onto $\mathbf{v}$ is defined as
>
> $$
>   a = \frac{1}{\|\mathbf{v}\|} \langle \mathbf{u}, \mathbf{v} \rangle,
> $$
>
> for all $\mathbf{u} \in V$ and $\mathbf{v} \in V \backslash \{\mathbf{0}\}$. 
>
> The **vector projection** $p$ of $\mathbf{u}$ onto $\mathbf{v}$ is defined as
>
> $$
> \mathbf{p} = a \bigg(\frac{1}{\|\mathbf{v}\|} \mathbf{v}\bigg) = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v},
> $$
>
> for all $\mathbf{u} \in V$ and $\mathbf{v} \in V \backslash \{\mathbf{0}\}$. 

It may be observed that $\mathbf{u} - \mathbf{p}$ and $\mathbf{p}$ are orthogonal since $\langle \mathbf{p}, \mathbf{p} \rangle = a^2$ and $\langle \mathbf{u}, \mathbf{p} \rangle = a^2$ which implies 

$$
    \langle \mathbf{u} - \mathbf{p}, \mathbf{p} \rangle = \langle \mathbf{u}, \mathbf{p} \rangle - \langle \mathbf{p}, \mathbf{p} \rangle = a^2 - a^2 = 0.
$$

Additionally, it may be observed that $\mathbf{u} = \mathbf{p}$ if and only if $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$; $\mathbf{u} = b \mathbf{v}$ for some $b \in \mathbb{K}$. Since 

$$
    \mathbf{p} = \frac{\langle b \mathbf{v}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} = b \mathbf{v} = \mathbf{u}.
$$

> *Theorem 2*: let $V$ be an inner product space then
>
> $$
>   | \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \|,
> $$
>
> is true for all $\mathbf{u}, \mathbf{v} \in V$. With equality only holding if and only if $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent.

??? note "*Proof*:"

    let $V$ be an inner product space and let $\mathbf{u}, \mathbf{v} \in V$. If $\mathbf{v} = \mathbf{0}$, then

    $$
        | \langle \mathbf{u}, \mathbf{v} \rangle | = 0 = \| \mathbf{u} \| \| \mathbf{v} \|,
    $$

    If $\mathbf{v} \neq \mathbf{0}$, then let $\mathbf{p}$ be the vector projection of $\mathbf{u}$ onto $\mathbf{v}$. Since $\mathbf{p}$ is orthogonal to $\mathbf{u} - \mathbf{p}$ it follows that

    $$
        \| \mathbf{p} \|^2 + \| \mathbf{u} - \mathbf{p} \|^2 = \| \mathbf{u} \|^2,
    $$

    thus 

    $$
        \frac{1}{\|\mathbf{v}\|^2} \langle \mathbf{u}, \mathbf{v} \rangle^2 = \| \mathbf{p}\|^2 = \| \mathbf{u} \|^2 - \| \mathbf{u} - \mathbf{p} \|^2,
    $$

    and hence

    $$
        \langle \mathbf{u}, \mathbf{v} \rangle^2 = \|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - \|\mathbf{u} - \mathbf{p}\|^2 \|\mathbf{v}\|^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,
    $$

    therefore

    $$
        | \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \|.
    $$

    Equality holds if and only if $\mathbf{u} = \mathbf{p}$. From the above observations, this condition may be restated to linear dependence of $\mathbf{u}$ and $\mathbf{v}$. 

A consequence of the Cauchy-Schwarz inequality is that if $\mathbf{u}$ and $\mathbf{v}$ aer nonzero vectors in an inner product space then

$$
    -1 \leq \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|} \leq 1,
$$

and hence there is a unique angle $\theta \in [0, \pi]$ such that 

$$
    \cos \theta = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|}.
$$

## Normed spaces

> *Definition 5*: a vector space $V$ is said to be a **normed linear space** if to each vector $\mathbf{v} \in V$ there is associated a real number $\| \mathbf{v} \|$ satisfying the following conditions
>
> 1. $\|\mathbf{v}\| > 0, \text{ for } \mathbf{v} \in V\backslash\{\mathbf{0}\} \text{ and } \| \mathbf{v} \| = 0, \text{ for } \mathbf{v} = \mathbf{0}$,
> 2. $\|a \mathbf{v}\| = |a| \|\mathbf{v}\|, \; \forall \mathbf{v} \in V \text{ and } a \in \mathbb{K}$,
> 3. $\| \mathbf{v} + \mathbf{w}\| \geq \|\mathbf{v}\| + \| \mathbf{w}\|, \; \forall \mathbf{v}, \mathbf{w} \in V$,
>
> is called the **norm** of $\mathbf{v}$. 

With the third condition, the *triangle inequality*. 

> *Theorem 3*: let $V$ be an inner product space then
>
> $$
>   \| \mathbf{v} \| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle},
> $$
>
> for all $\mathbf{v} \in V$ defines a norm on $V$. 

??? note "*Proof*:"

    Will be added later.

We therefore have that the Euclidean length (definition 2) is a norm, justifying the notation.