Parseval's Identity in Finite Dimensions

Norms and Orthonormal-Basis Coordinates

An orthonormal basis gives each vector a list of scalar coordinates, one for each basis vector. Parseval’s identity says that the squared norm of the vector can be recovered exactly from that list: add the squared absolute values of the coordinates. This matters because it connects a geometric quantity, the norm, to numerical data in a basis. The statement applies to real and complex inner product spaces, but in the complex case the absolute value is essential.

Let \(V\) be a finite-dimensional inner product space over \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). Use the convention that the inner product is linear in its first slot and conjugate-linear in its second. Thus, for complex scalars \(a,b\), \(\langle a\mathbf{u},b\mathbf{v}\rangle=a\overline{b}\langle\mathbf{u},\mathbf{v}\rangle\). In the real case, conjugation has no effect. The norm is defined by \(\|\mathbf{v}\|=\sqrt{\langle\mathbf{v},\mathbf{v}\rangle}\), so \(\|\mathbf{v}\|^2=\langle\mathbf{v},\mathbf{v}\rangle\).

Because the list is a basis, every vector has exactly one such expansion. Orthonormality lets us extract each coefficient using one inner product. We record that fact before proving the identity.

Proof. Fix an index \(j\). By linearity in the first slot and orthonormality,

Every term in the sum is zero except the term with \(k=j\), whose inner product factor is \(1\). This proves the claim for each coordinate. \(\square\)

Parseval’s Identity

Proof. Write the basis expansion \(\mathbf{v}=\sum_{j=1}^{n}c_j\mathbf{b}_j\). By the definition of the norm and the two-slot rules for the inner product,

For the third equality, orthonormality makes every term with \(j\ne k\) equal to zero, while \(\langle\mathbf{b}_j,\mathbf{b}_j\rangle=1\). In the real case \(\overline{c_j}=c_j\), so the same calculation gives \(\sum_j c_j^2\). If \(V\) has dimension zero, its only vector is \(\mathbf{0}\); the sum over the empty basis is \(0\), and the identity still holds. \(\square\)

The finite-dimensional hypothesis ensures that the basis expansion and sum have finitely many terms. The theorem does not say that the sum of the coordinates themselves equals the norm, nor that the norm equals the sum of their absolute values. It is specifically the squares of the absolute values that add to the squared norm.

Worked Calculations in Real Spaces

Why Complex Coordinates Need Absolute Squares

Over \(\mathbb{C}\), the coordinate \(c_j\) can have an imaginary part. Its square \(c_j^2\) can be complex or negative, whereas its squared absolute value \(c_j\overline{c_j}\) is a nonnegative real number. Since a squared norm is a nonnegative real number, Parseval’s identity uses \(|c_j|^2\). The inner product convention also affects the coefficient formula: with linearity in the first slot, \(c_j=\langle\mathbf{v},\mathbf{b}_j\rangle\), as proved above.

Coordinates Preserve Norm

Parseval’s identity has a useful interpretation: the coordinate list has exactly the same norm as the original vector. For a coordinate vector \(\mathbf{c}=[\mathbf{v}]_{\mathcal{B}}\in\mathbb{F}^n\), its usual Euclidean norm satisfies \(\|\mathbf{c}\|^2=\sum_{j=1}^{n}|c_j|^2\). Therefore Parseval gives \(\|\mathbf{v}\|^2=\|[\mathbf{v}]_{\mathcal{B}}\|^2\), and since norms are nonnegative, \(\|\mathbf{v}\|=\|[\mathbf{v}]_{\mathcal{B}}\|\).

Proof. By uniqueness of basis expansions, subtracting the expansion of \(\mathbf{w}\) from the expansion of \(\mathbf{v}\) gives \([\mathbf{v}-\mathbf{w}]_{\mathcal{B}}=[\mathbf{v}]_{\mathcal{B}}-[\mathbf{w}]_{\mathcal{B}}\). Apply Parseval’s identity to \(\mathbf{v}-\mathbf{w}\). Its \(j\)th coordinate is the difference of the \(j\)th coordinates of \(\mathbf{v}\) and \(\mathbf{w}\), which yields the displayed sum. \(\square\)

This is why orthonormal coordinates are especially convenient for comparing vectors: distances can be calculated using coordinate differences without first returning to the original representation. The conclusion depends on the basis being orthonormal. With a general basis, the coordinate squares do not in general add up to the original squared norm.

Common Pitfall: Squaring Complex Coordinates

A tempting but incorrect complex calculation is to add \(c_j^2\) instead of \(|c_j|^2\). In the \(\mathbb{C}^2\) example, the coordinates are \(c_1=(1-i)/\sqrt{2}\) and \(c_2=(3-i)/\sqrt{2}\). Using ordinary squares gives

which is not the squared norm \(6\). The error is that ordinary squaring does not pair each coordinate with its complex conjugate. The correction is

In a real inner product space, the coordinates equal their own conjugates, so writing \(c_j^2\) is correct there. In a complex inner product space, keep the absolute-value notation or explicitly write \(c_j\overline{c_j}\). This small distinction is exactly what makes the right-hand side real and nonnegative.