If I have a set of Linear Independent vectors $\{v_i\}^d_{i=1}$ that span a space V, (they are not orthogonal in general). How can one construct a set of vectors $\{w_i\}^d_{i=1}$ that fulfills, $$\langle v_i,w_j\rangle=\delta_{ij}$$ That involves the inner product.
In a book I've seen a formula that involves the determinant. But is not explained how to obtain it, if someone can give me some help with that.
Lets denote by $B=\left(b_{ij}\right)$ the matrix with components $$b_{ij}:=\left\langle v_i,v_j\right\rangle.$$ Also, let's assume such $w_j$ as in your question exist. Let $A=\left(a_{ij}\right)$ be the matrix that satisfies $$w_j=\sum_k a_{jk}v_k.$$ Such $a_{jk}$'s must exist since the $v_j$ form a basis of $V$. Now, we want: $$\delta_{ij}=\langle v_i,w_j\rangle=\sum_k a_{jk}\langle v_i,v_k\rangle=\sum_ka_{jk}b_{ki}.$$ Thus, we find $$A=B^{-1}$$ from which we get the $w_j$'s easily via the formula provided above.