## Proof That a Linear Operatior on a Vector Space to the Set of Real Numbers is the Inner Product

Let
$\mathbb{R}^n$
be the vector space of real n-tuples of the form
$(x_1,x_2,...,x_n)$
.
Let the inner product
$\phi$
on this vector space satisfy
$(x_1,x_2,...,x_n) \cdot (y_1,y_2,...,y_n)=x_1y_1+x_2y_2+...+x_ny_n$

The inner product defined above is a linear operator from
$\mathbb{R}^n$
to
$\mathbb{R}$
.
The inner product above is linear in both arguments (is bilinear) and satisfies
$\phi(a \vec{x}_1+b \vec{x}_2, \vec{y})=(a \vec{x}_1+b \vec{x_2}) \cdot \vec{y}=a \vec{x_1} \cdot \vec{y}+ b \vec{x}_2 \cdot \vec{y}$

$\phi(\vec{x},a \vec{y}_1+b \vec{y_2}, )= \vec{x} \cdot (a \vec{y}_1+b \vec{y}_2)=a \vec{x} \cdot \vec{y_1}+ b \vec{x} \cdot \vec{y}_2$

To show that
$\phi$
is the only inner product, let
$\vec{e}_1=(1,0,0,...,0), \vec{e}_2=(0,1,0,0,...,0),..., \vec{e}_n=(0,0,0,0,...,1)$
.
Then if
$\vec{x} in \mathbb{R}^n$
we can write
$\vec{x}=\sum_{i=1}^n x_i \vec{e}_i$
where the
$x_i$
are well defined. Applying
$\phi$
to
$\vec{x}$
gives
$\phi(\vec{x})= \sum_{i=1}^nx_i \phi(\vec{e}_i)$

Hence the values of
$\phi$
on a basis for
$\mathbb{R}^n$
determines
$\phi$
uniquely. These values are scalars and are the coordinates of some vector
$\vec{y}$
in
$\mathbb{R}^n$
, and we can write
$\vec{y}= \sum_{i=1}^n \phi ({e}_i) \vec{e}_i$
.
Then
$\phi (\vec{x}_)= ( \vec{x} , \vec{y} >$
.
Then the only function from a vector space to the set of real numbers is the inner product above.