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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.

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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 $y\in H$.

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  • $\begingroup$ Excelent! Totally Works! $\endgroup$
    – Tom Builder
    Commented Jun 5, 2015 at 0:03
  • $\begingroup$ I just have a question, why does the second term in the definition of the inner product have to be? I think it still works even without that! $\endgroup$
    – Tom Builder
    Commented Jun 5, 2015 at 0:18
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    $\begingroup$ @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 $H$. $\endgroup$
    – Disintegrating By Parts
    Commented Jun 5, 2015 at 0:58

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