Let $V$ be a real inner product space with basis $U=(v_1, \ldots, v_n)$ and $A \in Mat_{n,n}(\mathbb R)$ with $(i,j)$-entry equal to $\langle v_i v_j \rangle$.
I've shown the identity $\langle x, y \rangle = [x]_U^T \cdot A \cdot [y]_U \ \forall x,y \in V$ by writing $x= c_1 v_1 + \ldots c_n x_n$ and $y= d_1 v_1 + \ldots d_n v_n$, and then expanding inner product using axioms and afterwards computing the matrix product.
However I must show $A$ is invertible.
I try to show this by writing $Ax = 0$ and show $x = 0$.
$Ax = 0 \iff x_1 \langle v_i, v_1 \rangle + \ldots + x_j \langle v_i, v_j \rangle + \ldots \langle v_i, v_n \rangle = 0 \iff \langle v_i, x_1 v_1 + \ldots + x_j v_j + \ldots + x_n v_n \rangle = 0$
I want to conclude $x_1 v_1 + \ldots + x_j v_j + \ldots + x_n v_n = 0$ and then since $v_i : 1 \le i \le n$ is a basis we have $x_j = 0 : 1 \le j \le n$. However, how can I do this ?
By the way, the matrix $A$ is called the Gram matrix for $v_1, \ldots, v_n$.