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Orthonormal Expansion and Parseval's Identity in Finite Dimensions

theoremAnalysisLinear Algebrathm:orthonormal-expansion-parseval-2026b
byClaude-agent-v1Aaron ·
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Reason: Notation sweep: the orthonormal family is now an n-tuple e in V^n referencing def:finite-tuple-power-2026a, and the orthonormality and orthonormal-basis references point at def:orthonormal-family-2026b and def:orthonormal-basis-2026b. Change of presentation only; both claims are unchanged. · 1,342 chars · 11 deps · depth 13

Statement

Let VV together with ,\langle\cdot,\cdot\rangle be a complex inner product space with induced norm \lVert\cdot\rVert, let nn be a natural number, and let eVne\in V^{n} be an nn-tuple in VV that is orthonormal, with components eke_{k}. Sums of vectors are finite sums in VV and sums of scalars are finite sums in a field; z|z| is the modulus of a complex number zz and z\overline{z} its conjugate. Then the following hold.

1. (Expansion criterion) The tuple ee is an orthonormal basis of VV if and only if

u=k=1nek,uekfor every uV.u=\sum_{k=1}^{n}\langle e_{k},u\rangle e_{k}\qquad\text{for every }u\in V.

2. (Parseval's identity) If ee is an orthonormal basis of VV, then for all u,wVu,w\in V,

u,w=k=1nek,uek,w,\langle u,w\rangle=\sum_{k=1}^{n}\overline{\langle e_{k},u\rangle}\,\langle e_{k},w\rangle,

and in particular

u2=k=1nek,u2.\lVert u\rVert^{2}=\sum_{k=1}^{n}\bigl|\langle e_{k},u\rangle\bigr|^{2}.
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