Discussion
An infinite dimensional, separable complex Hilbert space is congruent to :
Answer: B
It is congruent to \(l_{2}\).
An infinite dimensional separable Hilbert space has a countable orthonormal basis \(\{e_{n}\}\), obtained by Gram-Schmidt from a countable dense set. The map \(x \mapsto (\langle x, e_{n}\rangle)_{n}\) sends it onto the square-summable sequences, and Parseval's identity \(\|x\|^{2} = \sum |\langle x, e_{n}\rangle|^{2}\) shows that the map preserves the inner product. So the space is isometrically isomorphic to \(l_{2}\).
\(l_{1}\), \(l_{3}\) and \(l_{\infty}\) are Banach spaces whose norms do not come from an inner product (they fail the parallelogram law), so no Hilbert space can be congruent to them.
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