Questions about Gram matrix When all the vectors are chosen from $R^n$, it is clear to prove Gram matrix invertible if and only if the set of vectors is linearly independent as shown below.
Gram matrix invertible iff set of vectors linearly independent
I think the trick is that $G$ can be written as $A^TA$, where $A$ is $(\mu_1,\cdots,\mu_n)$, $\mu_i$ is k dimensional column vector. But if the set of vector is not belong to $R^n$, how can I prove it.
Here is something I found, but can not clearly understand. $G_{ij}=(\mu_i,\mu_j)$, $\mu_i$ belongs to a inner product space $X$.
$G$ is invertible iff only $0$ is the solution to
$(\sum_{j=1}^{n} a_j \mu_j,\mu_k)=\sum_{j=1}^{n}(\mu_j,\mu_k)a_j=0$
$\sum_{j=1}^{n} a_j \mu_j=0$
$\iff (\sum_{j=1}^{n} a_j \mu_j,\sum_{j=1}^{n} a_j \mu_j)=0$
$\iff (\sum_{j=1}^{n} a_j \mu_j,\mu_k)=\sum_{j=1}^{n}(\mu_j,\mu_k)a_j=0$
What does the proof mean? How can I get my questions solved. Appreciate for your helping hands.
 A: I think this question is very elementry in linear algebra, and there are lots of viewpoints to understand this. The "only if" part is trivial, so I only show the "if" part.

First and to me the most standard way is SVD. Firstly, since the columns of $A$ are independent, $A$ is m-by-n matrix with $m\ge n$. Write $A=U(\begin{array}{c}\Sigma\\0\end{array})V^T$ as the SVD of $A$, where $\Sigma=\text{diag}(\sigma_1, \ldots, \sigma_n)$ with $\sigma_i\neq 0$ for all $i$, and $U$ and $V$ are orthogonal matrices. Then
$$A^T A = V\Sigma^2V^T$$
is clearly nonsingular.

Second, we can think in terms of linear mapping. Let $V=span(v_1, \ldots, v_n)$ by any subspace of $\mathbb{R}^n$ with unit-length basis $v_1,\ldots, v_n$. We need to show that if $x\in V,x\neq 0$, then there exists $v_i$ such that $x\cdot v_i \neq 0$. Indeed,
$$x = (x\cdot v_1) v_1 + \cdots (x\cdot v_n) v_n$$
$x\neq 0$, there exists $x\cdot v_i$ that is not zero.
Now Let $A=(a_1, \cdots, a_n)$. Suppose $x\in Range(A)$, $A^Tx$ is the coordinate of $x$ under the basis $a_1,\ldots, a_n$. Therefore, the mapping $f:Range(A) \rightarrow \mathbb{R}^m, f(x)=A^Tx$ is injective. Now let $g:\mathbb{R}^m\rightarrow Range(A), g(x) = Ax$, since $a_1,\ldots, a_n$ are independent, $g$ is injective. Therefore, the composite $h=f\circ g:\mathbb{R}^m\rightarrow \mathbb{R}^m, h(x) = A^TAx$ is injective. Since $m=m$, $h$ is also surjective. Hence $h$ is invertible, and so is the matrix $A^TA$.

