Let $\{r_i\}$ be a finite list of independent vectors. Denote the inner product of two vectors as $\langle r_i,r_j \rangle$. Define a matrix $V$ such that its $(i,j)$-th entry is $\langle r_i,r_j \rangle$. Show that $V$ is invertible.
I came up with this problem myself while studying projections.
My attempt is as follows:
If $V$ is not invertible, we may write a row (assume its $k$-th row) of $V$ as linear combination of some other rows :
$$(\langle r_k,r_1 \rangle,\langle r_k,r_2 \rangle,\dots)=a \cdot (\langle r_{l_1},r_1 \rangle,\langle r_{l_1},r_2 \rangle,\dots)+b\cdot(\langle r_{l_2},r_1 \rangle,\langle r_{l_2},r_2 \rangle,\dots)+\cdots$$
$$\Rightarrow \quad z:=(r_k-ar_{l_1}-br_{l_2}-\cdots)\perp r_j\ , \quad \forall j$$
by entry-wise comparison where $a,b,…\neq 0$.
So $z \in \text{span}(r_1,…)^\perp$ but $z \in \text{span}(r_1,…)$. This implies $z=0$,
$$\implies r_k=ar_{l_1}+br_{l_2}+\cdots$$
contradicting independence of $\{r_i\}$.
Best Answer
Your proof is correct. It is notable that your matrix is sometimes called a Gramian matrix. The wikipedia page provides an alternative proof of your fact, which you might want to take a look at.