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Let $x_i$ be a basis of Hilbert space $X$ (NOT necessarily orthogonal)

How do I show that $\text{det}((x_i,x_j)_H)_{ij} \neq 0$ for $i,j=1,...,n$?

I see this fact used in Galerkin approximation solution.

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To me, this is a consequence of the fact that the functionals defined by $x \mapsto \langle x_i, x \rangle$ are linearly independent. –  Tunococ Aug 30 '13 at 13:10
    
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2 Answers

up vote 1 down vote accepted

Hints:

Let us define the row vectors $\;r_m:=\left((x_m,x_1)\,,\,(x_m,x_2),\ldots,(x_m,x_n)\right)$ :

$$\begin{align*}\bullet&\;\det (x_i,x_j)=0\iff \exists\;1\le k\le n\;\;s.t.\;\;r_k=\sum_{i=1}^{k-1}a_ir_i\;,\;a_i\in\Bbb F(=\Bbb R,\Bbb C)\\ \bullet&\;\text{The above means that}\;\;\forall\;1\le j\le n\;,\;(x_k,x_j)=\sum_{i=1}^{k-1}a_i(x_i,x_j)\implies\\ &\implies\left(x_k-\sum_{i=1}^{k-1}a_ix_i\,,\,x_j\right)=0\implies\;x_k-\sum_{i=1}^{k-1}a_ix_i\perp x_j\\ \bullet&\text{What's the only vector perpendicular to the whole space? What's the contradiction?}\end{align*}$$

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Using the language of differential geometry, $$\det(\langle x_i,x_j\rangle)=\langle x_1\wedge x_2\wedge...\wedge x_n, x_1\wedge x_2\wedge...\wedge x_n\rangle=|| x_1\wedge x_2\wedge...\wedge x_n||~^2\neq0$$ since $\{x_i\}_{i=1..n}$ is basis for space.

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