Exercise 1750Linear algebra › Inner product spacesL1Added 3 Nov 2026

Parseval's identity

#Parseval identity#orthonormal basis#norm

Statement

Let $(e_1,\ldots,e_n)$ be an orthonormal basis of a finite-dimensional inner product space $E$ . Show that for every $v\in E$ ,
$\|v\|^2=\sum_{i=1}^n|\langle v,e_i\rangle|^2.$

Hint

Use Bessel's inequality and the fact that $v=\sum_i\langle v,e_i\rangle e_i$ (since the basis is complete).

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Solution

Expansion. Since $(e_i)$ is a basis, $v=\sum_i c_i e_i$ with $c_i=\langle v,e_i\rangle$ (because $\langle v,e_j\rangle=\sum_i c_i\langle e_i,e_j\rangle=c_j$ ).

Norm. $\|v\|^2=\langle\sum_i c_i e_i,\sum_j c_j e_j\rangle=\sum_i|c_i|^2$ by orthonormality.

Conclusion. $\|v\|^2=\sum_i|\langle v,e_i\rangle|^2$ = Parseval's identity.

Remark. This is the equality case of Bessel's inequality when the orthonormal family is complete (a basis).

Parseval's identity

Statement

Solution

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