# Find an inner product that makes a given set of linearly independent vectors orthogonal

I need to find an inner product such that given a set $S$ of linearly independent vectors in a Hilbert space $H$, $S$ will be orthogonal with these product.

I thought Gram -Schmidt Process would help but it's not, because for the process you already have the inner product.

Start with a set $\{ x_1,\cdots,x_n\}$. Define $$U : \mathbb{C}^{n} \rightarrow H$$ by $$U(\alpha_1,\cdots,\alpha_n) = \alpha_1 x_1+\cdots+\alpha_n x_n$$ Let $P$ the be orthogonal projection of $H$ onto the closed subspace spanned by $\{x_1,\cdots,x_n\}$. Define $$(x,y)_{\mbox{new}} = (U^{-1}Px,U^{-1}Py)_{\mathbb{C}^{n}}+((I-P)x,(I-P)y)_{H}.$$ Because $(I-P)x_k=0$ for $1 \le k \le n$ and $U^{-1}x_k$ is the $k$-th standard basis element in $\mathbb{C}^{n}$, then $(x_j,x_k)_{\mbox{new}}=\delta_{j,k}$ and $(x_j,(I-P)y)_{\mbox{new}}=0$ for all $1 \le j \le n$ and $y\in H$.
• @TomBuilder : The second part extends the inner product to all of $H$ in such a way that the vectors $x_1,\cdots,x_n$ are orthogonal to the original orthogonal completment of the subspace generate by the $x_k$. Without the second term, $(\cdot,\cdot)_{\mbox{new}}$ would not be positive definite. By the way, the original inner product and this one generate the same topology on $H$. – DisintegratingByParts Jun 5 '15 at 0:58