Suppose $|v_1\rangle, |v_2\rangle ,\cdots |v_k\rangle \in S_N$ and suppose a matrix $G$ form by the inner product of these vectors $$G_{ij}=\langle v_i|v_j\rangle $$ I'm trying to prove that $\text{det} (G)\geq 0$ where equality hold for if vectors are linearly dependent.
Since $$\langle v_i|v_j\rangle =\langle v_j|v_i\rangle ^*\rightarrow G_{ij}=G_{ji}^*\rightarrow G=G^\dagger$$ this means that eigenvalues will be positive. In the diagonal form, $$G=\text{diag}(g_1,g_2,\cdots ,g_k)$$ The determinant can be written as $$\text{det}(G)=\prod_ig_i$$ To prove this is positive, We can prove that all the eigenvalues are positive but that is much more restrictive. I'm not sure, How do I proceed? Please help me with this.
This matrix is the Gramian matrix corresponding to $v_1, \ldots, v_k$. It's clearly Hermitian, and you can show that it is positive-semidefinite by computing $x^\dagger Gx$ and showing it is always non-negative. Indeed, we have \begin{align*} Gx &= \begin{bmatrix} \langle v_1 \mid v_1 \rangle & \langle v_1 \mid v_2\rangle & \cdots & \langle v_1 \mid v_k\rangle \\ \langle v_2 \mid v_1 \rangle & \langle v_2 \mid v_2\rangle & \cdots & \langle v_2 \mid v_k\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle v_k \mid v_1 \rangle & \langle v_k \mid v_2\rangle & \cdots & \langle v_k \mid v_k\rangle \end{bmatrix}\begin{bmatrix} x_1 \\x_2 \\ \vdots \\ x_k \end{bmatrix} \\ &= \begin{bmatrix} x_1\langle v_1 \mid v_1 \rangle + x_2\langle v_1 \mid v_2\rangle + \cdots + x_k\langle v_1 \mid v_k\rangle \\ x_1\langle v_2 \mid v_1 \rangle + x_2\langle v_2 \mid v_2\rangle + \cdots + x_k\langle v_2 \mid v_k\rangle \\ \vdots \\ x_1\langle v_k \mid v_1 \rangle + x_2\langle v_k \mid v_2\rangle + \cdots + x_k\langle v_k \mid v_k\rangle \end{bmatrix} \\ &= \begin{bmatrix} \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_1 \rangle \\ \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_2 \rangle \\ \vdots \\ \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_k \rangle \end{bmatrix} \end{align*} Then, \begin{align*} x^\dagger Gx &= \begin{bmatrix} \overline{x}_1 & \overline{x}_2 & \cdots & \overline{x}_k\end{bmatrix} \begin{bmatrix} \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_1 \rangle \\ \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_2 \rangle \\ \vdots \\ \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_k \rangle \end{bmatrix} \\ &= \overline{x}_1\langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_1 \rangle + \overline{x}_2\langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_2 \rangle \\ &+ \cdots + \overline{x}_k\langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid v_k \rangle \\ &= \langle x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \mid x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \rangle \\ &= \|x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \|^2 \ge 0. \end{align*} This implies that all the eigenvalues are non-negative. Note also that, if $v_1, \ldots, v_k$ are linearly independent, then for non-zero $x$, $$x^\dagger G x = \|x_1 v_1 + x_2 v_2 + \ldots + x_k v_k \|^2 > 0.$$ This makes $G$ positive-definite, and hence all its eigenvalues are strictly positive.