Representing inner products on $\mathbb{R^n}$ using matrices I would like to ask two basic questions about inner products on $\mathbb{R^n}$:
Given an inner product $\langle u, v\rangle$ on $\mathbb{R^n}$,
1) Am I correct in thinking that the matrix $A$ with $a_{ij}=\langle e_j,e_i\rangle$ satisfies $\langle u, v\rangle=\big(Au\big)\cdot v$ 
$\;\;$for all $u,v\in \mathbb{R^n}$ where $e_1,\cdots, e_n$ are the standard basis vectors?
2) When is it possible to find an invertible matrix $C$ with $\langle u,v\rangle=\big(Cu\big)\cdot\big(Cv)$ for all $u,v\in \mathbb{R^n}$?
 A: 1) Yes, see the example below (equation (1)). An inner product is always associated with a positive definite (p.d.) matrix $A$.
2) The existence of such a matrix $C$ is well known; one can even give it an upper triangular form ; it is called the Cholesky matrix, and the decomposition $A=C^TC$ is called the Cholesky decomposition of matrix $A$:  https://en.wikipedia.org/wiki/Cholesky_decomposition
Edit : Here is an example of a p.d. matrix $A$ and its Cholesky decomposition
$$A=\begin{bmatrix}1&1&1\\1&2&3\\1&3&6\end{bmatrix}=
\begin{bmatrix}1&0&0\\1&1&0\\1&2&1\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}$$
An important point, not always underlined, is that Cholesky decomposition is equivalent to the (Gauss) decomposition of a p.d. quadratic form into the sum of squares of independant linear forms. In the case of the quadratic form associated with matrix $A$, i.e.,
$$Q(x,y,z)=x^2+2y^2+6z^2+2xy+2xz+6yz=\begin{bmatrix}x&y&z\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&3\\1&3&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} \ \ \ (1)$$
The algorithm is as follows: rank the variables (here, there is a natural order x, y z). Consider the first variable $x$ and put the polynomial in canonical form with respect to this variable, i.e., here:
$$Q(x,y,z)=(x+(y+z))^2+R(x,y)$$
where $R(x,y)$ is a polynomial in the remaining variables, and begin back the same operation on this new polynomial, this time with variable $y$, etc., ending with the final decomposition
$$Q(x,y,z)=(x+y+z)^2+(y+2z)^2+z^2$$
which can be written under the matrix form:
$$\begin{bmatrix}(x+y+z)&(y+2z)&z\end{bmatrix}
\begin{bmatrix}(x+y+z)\\(y+2z)\\z\end{bmatrix}=\begin{bmatrix}x&y&z\end{bmatrix}\begin{bmatrix}1&0&0\\1&1&0\\1&2&1\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} \ (2)$$
Comparison between (1) and (2) does not need commentaries.
