\[\mathbb{R}^n\]
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.