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.