Gram's matrix demonstration Given the Gram's matrix, A, where $A_{ij}=(\psi_{i}|\psi_{j})$ (of course $\psi_{i}$ $\in V$), can someone prove that
$$
\textrm{If} \hspace{2mm} (\psi_{i}|A|\psi_{i})>0 \hspace{5mm} (\textrm{i.e A is positively defined}) \implies \{\psi_{i}\}_{i=1}^{i=N} \hspace{5mm} \textrm{form a a set of independent vectors ( i.e} \hspace{2mm} (\psi_{i}|\psi_{j})=0)
$$ 
? 
And in this case, show that conversely all Gram matrix of a set of vectors is a semipositively defined operator.

*

*If A is positively defined then it is nonsingular. Lets assume that $\psi_{1} = \lambda_2\psi_{2}+\dots + \lambda_n\psi_{n}$ so that our vectors are linearly dependent. Then the first row (or column) of $A$ is a linear combination of other rows (columns) with coefficients $\lambda_2, \dots, \lambda_n$, so that $A$ is singular. Contradiction.

*A Gram matrix of a set of vectors forming matrix $M$ is $G=X^TX$. So that $w^TGw = w^TX^TXw = (Xw)^TXw = \|Xw\|_2^2 \ge 0$.

