> *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
> 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
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
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.
> 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
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
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
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
> *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. $\|a \mathbf{v}\| = |a| \|\mathbf{v}\|, \; \forall \mathbf{v} \in V \text{ and } a \in \mathbb{K}$,