I am hoping someone can corroborate my chain of reasoning. Let $P=[p_{ij}] \in M_n(\Bbb{C})$ be a gram matrix. The there exists a inner product space $(V,\langle, \rangle_V)$, which is possibly infinite dimensional, and vectors $v_1,...,v_n \in V$ such that $p_{ij} = \langle v_i, v_j \rangle_V$. Since all Gram matrices are PSD (and vice-versa), there exists a matrix $C \in M_n(\Bbb{C})$ such that $P = C^*C$. Let the columns of $C$ be denoted by $w_1,...,w_n$, which are vectors in $\Bbb{C}^n$. Then it's clear that $\langle w_i, w_j \rangle_{\Bbb{C}^n} = \langle v_i, v_j \rangle_V$ for all $i$ and $j$ , unless I incorrectly defined and indexed the columns of $C$ (hopefully someone will point that out). Why it is clear can be seen by computing the $(i,j)$-th entry of $C^*C$ and noting that it is the inner product the $i$-th row of $C^*$ with the $j$-th column of $C$ (or maybe the other way around; I always mess this up).
Thus, the entire collection of Gram matrices can be generated by vectors in finite dimensional, meaning that any Gram matrix generated by an infinite dimensional inner product space can be likewise generated by a finite dimensional one.
Is there another way of arriving at this fact? I feel like the above method is a tad circular--just a tad, not entirely circular. I was thinking of something like the following. Let $v_1,...,v_n \in V$ be, WLOG, linearly independent vectors, which implies $\dim (span \{v_1,...,v_n\}) = n$. Letting $W := span( \{v_1,...,v_n\})$, there exists an isomorphism $f : W \to \Bbb{C}^n$, which is continuous since it is between finite dimensional vector spaces, and $\{f(v_1),...,f(v_n)\}$ serves as a basis for $\Bbb{C}^n$. Unfortunately, $f$ need not be a isometry, so there isn't a quick kill. How do we finish the proof?
