Any inner product can be defined through the standard product and a Matrix $(\cdot | \cdot )_0 $ an inner Product in $\mathbb{R}^n$.
Show that there is one and only one symmetric, positive Definite $n \times n$ matrix $A$ such that:
$$ (x |y)_0 = (Ax|y) $$ for all vectors in $\mathbb{R}^n$. And $(x|y)$ the standard inner product.
So i have shown that if such A exists then it must be unique, that part is easy. But I have troubles showing the existence.
I thought it could be possible, since it is possible to reach any linear combination of the coordinates of x and y with a proper matrix, but I don't know how to write that mathematically and if it is enough...
 A: Now first of all, lets use proper notation! 
what we want is a matrix s.t $xAy^*=\langle x , y \rangle$.
Since you already have an inner product, you have a basis (i.e. worst case svenario, pick a ONB(orthonormal basis))$\{b_1,...,b_n\}$. Now define $A_{i,j}:= \langle b_i , b_j \rangle$. then this matrix suffices the formula above. I leave the verification (using representation into basis elements) to you.
A: Let $\{e_1,\dots,e_n\}$ be the standard basis for $\Bbb R^n$. If such matrix $A$ exist, put $x=e_i$, $y=e_j$ and the right-hand-side would become $(Ae_i\vert e_j)=a_{ji}$, the $(j,i)$-entry of $A$. The left-hand-side shall be equal to this number, i.e. we have $a_{ji}=(e_i\vert e_j)_0$. This deals with uniqueness of $A$ actually.
You then show that this $A$ satisfies the equation for all $x,y$, because we only know the equality holds when $x,y$ are standard basis vectors. But, you don't quite need to prove this, because there is the following proposition:

Proposition: If two real-valued bilinear functions $B_1,B_2$ on $\Bbb R^n$ satisfy the property that there exists a basis $\{v_1,\dots,v_n\}$ for $\Bbb R^n$ such that $B_1(v_i,v_j)=B_2(v_i,v_j)$ for all $i,j$, then in fact $B_1=B_2$.

If you know how to prove this and how to apply this theorem, you are done.
If not, prove in the old-fashioned way: let $x=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix},y=\begin{bmatrix}y_1\\\vdots\\y_n\end{bmatrix}$, and prove the equality $(x\vert y)_0 = (Ax\vert y)$ with brute force.
