Orthonormal Bases and Coordinates

Suppose you want the coordinates of a vector relative to a basis of a five-dimensional space. In general this means solving a linear system. You write the vector as an unknown combination of basis vectors. Then you expand, and match coefficients. This can be genuinely painful. But suppose the basis vectors are pairwise orthogonal. Suppose also each has length one. Then the entire computation collapses into a single inner product per coordinate. No system to solve at all. This is the central payoff of working with an orthonormal basis. Once a basis is orthonormal, extracting coordinates becomes mechanical. The kk-th coordinate of a vector is simply its inner product with the kk-th basis vector. This formula is often called the Fourier coefficient formula. It is the linear-algebraic ancestor of the Fourier series formula from analysis. It explains why orthonormal bases are the preferred coordinate systems throughout mathematics, physics, and engineering. The idea matters because it converts two hard problems into easy, formulaic ones. Those problems are finding coordinates, and computing inner products or norms of vectors given in coordinates. Once coordinates relative to an orthonormal basis are known, the inner product of two vectors becomes the ordinary dot product of their coordinate tuples. The norm becomes the ordinary Euclidean length of the coordinate tuple. All of the geometric intuition of Euclidean space transports faithfully to any finite-dimensional inner product space. This holds provided one uses an orthonormal basis. This article proves the Fourier coefficient formula. It also proves the componentwise formulas for inner products and norms. It builds directly on Orthogonality and Orthogonal Sets and the Triangle Inequality and Norm Properties. It sets the stage for the Gram-Schmidt Process, which manufactures orthonormal bases out of arbitrary ones. It also prepares for Orthogonal Complements and the Bessel and Parseval Identities. These extend these ideas to infinite-dimensional and non-basis settings.

Key takeaways
  • An orthonormal basis is a basis whose vectors are pairwise orthogonal and each have norm 11.
  • If (e1,…,en)(e_1,\dots,e_n) is an orthonormal basis, every vector vv satisfies v=∑k⟨v,ek⟩ekv = \sum_k \langle v, e_k\rangle e_k.
  • The scalar ⟨v,ek⟩\langle v, e_k \rangle is called the kk-th Fourier coefficient of vv; it is exactly the kk-th coordinate.
  • Relative to an orthonormal basis, inner products become dot products of coordinate tuples: ⟨u,v⟩=∑kakbk‾\langle u,v\rangle = \sum_k a_k \overline{b_k}.
  • Norms become Euclidean lengths of coordinate tuples: ∥v∥2=∑k∣⟨v,ek⟩∣2\|v\|^2 = \sum_k |\langle v,e_k\rangle|^2.
  • Key idea: orthonormality turns "solve a linear system for coordinates" into "compute nn inner products."

Where this lives today

Orthonormal coordinates are the computational backbone of enormous swaths of applied mathematics. In numerical linear algebra, algorithms such as QR decomposition and the Gram-Schmidt Process exist precisely to convert an arbitrary basis into an orthonormal one, after which solving least-squares problems, computing projections, and inverting systems all become numerically stable, componentwise operations rather than ill-conditioned matrix inversions. In signal processing, discrete and continuous Fourier transforms, wavelet transforms, and the discrete cosine transform (which underlies JPEG image compression) are all instances of expanding a signal in an orthonormal basis and reading off Fourier coefficients via inner products — exactly the formula proved above, applied to function or sequence spaces. In quantum mechanics, physical states are vectors in a complex Hilbert space, and an orthonormal basis of eigenstates lets physicists compute transition amplitudes and probabilities as inner products of state vectors with basis vectors — the Fourier coefficients here have direct physical meaning as probability amplitudes. In statistics and machine learning, principal component analysis produces an orthonormal basis (the eigenvectors of a covariance matrix) in which data coordinates are uncorrelated, and dimensionality reduction relies on truncating the componentwise sum from part (c) of the theorem. In computer graphics and robotics, orthonormal frames (rotation matrices) are used constantly because their inverse is their transpose, which is a direct matrix restatement of the coordinate formula. In cryptography and coding theory, orthonormal (and more generally orthogonal) bases underlie lattice constructions and error-correcting code design, where controlling norms via componentwise sums of squares is essential to bounding error probabilities.

Ask the tutor: If I only know an orthogonal (not orthonormal) basis for a subspace, how do I adapt the Fourier coefficient formula to compute the orthogonal projection of a vector onto that subspace?

DefinitionOrthonormal basis and Fourier coefficients

Let VV be a finite-dimensional inner product space over F\mathbb{F} (where F=R\mathbb{F} = \mathbb{R} or C\mathbb{C}), with inner product ⟨⋅,⋅⟩\langle \cdot,\cdot\rangle conjugate-linear (or linear, if F=R\mathbb{F}=\mathbb{R}) in the second slot, and ∥v∥=⟨v,v⟩\|v\| = \sqrt{\langle v,v\rangle}. A list of vectors (e1,…,en)(e_1,\dots,e_n) in VV is called orthonormal if

⟨ei,ej⟩=δij={1i=j0i≠jfor all i,j∈{1,…,n}.\langle e_i, e_j \rangle = \delta_{ij} = \begin{cases} 1 & i=j \\ 0 & i \ne j\end{cases} \qquad \text{for all } i,j \in \{1,\dots,n\}.

An orthonormal basis of VV is an orthonormal list (e1,…,en)(e_1,\dots,e_n) that is also a basis of VV, i.e. n=dim⁡Vn = \dim V. Given an orthonormal basis (e1,…,en)(e_1,\dots,e_n) of VV and a vector v∈Vv \in V, the scalar

v^k:=⟨v,ek⟩\hat v_k := \langle v, e_k \rangle

is called the kk-th Fourier coefficient of vv relative to this basis.

Intuition

Think of an orthonormal basis as a set of perfectly perpendicular, unit-length measuring sticks. To find "how much" of e1e_1 is inside a vector vv, you do not need to disentangle contributions from e2,e3,…e_2, e_3, \dots at all — projecting vv onto the line through e1e_1 automatically ignores every other direction, precisely because they are orthogonal to e1e_1. The inner product ⟨v,ek⟩\langle v, e_k\rangle is that projection length (with sign, or phase in the complex case). Contrast this with a skewed, non-orthogonal basis: to read off the coordinate of vv along one axis, you must first account for how the axes lean into each other, which is exactly what solving a linear system (inverting a change-of-basis matrix) does. Orthonormality removes the leaning entirely: the "change of basis matrix" whose columns are the eke_k becomes an isometry (orthogonal or unitary), and its inverse is simply its conjugate transpose — so recovering coordinates is a matter of taking inner products, not inverting a matrix. The second piece of intuition is that once you have these coordinates, the abstract inner product space "looks like" Fn\mathbb{F}^n with the standard dot product. Lengths, angles, and orthogonality in VV are computed by ordinary Euclidean formulas applied to the coordinate tuples. The orthonormal basis is the dictionary that translates VV's geometry into the geometry you already know from Rn\mathbb{R}^n.

Figure
Reading off Fourier coefficients by orthogonal projection
e₁ e₂ v ⟨v, e₁⟩ = 3.0 ⟨v, e₂⟩ = 2.0 1 1 v = 3.0 e₁ + 2.0 e₂
TheoremFourier Coefficient Formula and Componentwise Inner Products

Let VV be a finite-dimensional inner product space over F∈{R,C}\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}, and let (e1,…,en)(e_1,\dots,e_n) be an orthonormal basis of VV. Then:

(a) Coordinate formula. For every v∈Vv \in V,

v=∑k=1n⟨v,ek⟩ ek.v = \sum_{k=1}^n \langle v, e_k \rangle\, e_k .

That is, the (unique) scalars a1,…,an∈Fa_1,\dots,a_n \in \mathbb{F} with v=∑kakekv = \sum_k a_k e_k are exactly ak=⟨v,ek⟩a_k = \langle v, e_k\rangle.

(b) Componentwise inner product. For all u,v∈Vu, v \in V, writing ak=⟨u,ek⟩a_k = \langle u, e_k\rangle and bk=⟨v,ek⟩b_k = \langle v, e_k \rangle,

⟨u,v⟩=∑k=1nakbk‾.\langle u, v \rangle = \sum_{k=1}^n a_k \overline{b_k}.

(c) Componentwise norm (Parseval-type identity for a basis). For every v∈Vv \in V,

∥v∥2=∑k=1n∣⟨v,ek⟩∣2.\|v\|^2 = \sum_{k=1}^n |\langle v, e_k\rangle|^2 .
Proof

We use the fact, following from linear independence of a basis, that every v∈Vv\in V has a unique representation v=∑k=1nakekv = \sum_{k=1}^n a_k e_k with ak∈Fa_k \in \mathbb{F}; this is simply the defining property of a basis and does not depend on orthonormality.

Proof of (a). Fix v∈Vv \in V and write v=∑k=1nakekv = \sum_{k=1}^n a_k e_k for the (unique) scalars given by the basis property. Fix an index j∈{1,…,n}j \in \{1,\dots,n\}. Using linearity of the inner product in the first slot and conjugate-linearity (or linearity, in the real case) in the second slot only as needed for scalars already in F\mathbb F, compute

⟨v,ej⟩=⟨∑k=1nakek, ej⟩=∑k=1nak⟨ek,ej⟩,\langle v, e_j \rangle = \Big\langle \sum_{k=1}^n a_k e_k,\ e_j \Big\rangle = \sum_{k=1}^n a_k \langle e_k, e_j \rangle,

where the first equality is the definition of vv and the second is linearity of the inner product in its first argument (an axiom of the inner product, hence not requiring further proof here). By orthonormality of (e1,…,en)(e_1,\dots,e_n) we have ⟨ek,ej⟩=δkj\langle e_k, e_j\rangle = \delta_{kj}, so every term in the sum vanishes except the k=jk=j term, which equals aj⋅1=aja_j \cdot 1 = a_j. Hence

⟨v,ej⟩=aj\langle v, e_j \rangle = a_j

for every j=1,…,nj = 1, \dots, n. Substituting these values of aja_j back into v=∑kakekv = \sum_k a_k e_k gives

v=∑k=1n⟨v,ek⟩ ek,v = \sum_{k=1}^n \langle v, e_k \rangle\, e_k,

which is the claimed formula. This also confirms that the abstract coefficients aka_k (guaranteed to exist and be unique by the basis property) coincide with the Fourier coefficients ⟨v,ek⟩\langle v, e_k\rangle.

Proof of (b). Let u=∑k=1nakeku = \sum_{k=1}^n a_k e_k and v=∑k=1nbkekv = \sum_{k=1}^n b_k e_k be the expansions from part (a), so ak=⟨u,ek⟩a_k = \langle u, e_k\rangle and bk=⟨v,ek⟩b_k = \langle v, e_k \rangle. Using linearity of the inner product in the first argument and conjugate-linearity in the second argument (the sesquilinearity axioms of the inner product; in the real case, ordinary bilinearity), we compute:

⟨u,v⟩=⟨∑i=1naiei, ∑j=1nbjej⟩=∑i=1n∑j=1naibj‾⟨ei,ej⟩.\langle u, v \rangle = \Big\langle \sum_{i=1}^n a_i e_i,\ \sum_{j=1}^n b_j e_j \Big\rangle = \sum_{i=1}^n \sum_{j=1}^n a_i \overline{b_j} \langle e_i, e_j \rangle.

By orthonormality, ⟨ei,ej⟩=δij\langle e_i, e_j \rangle = \delta_{ij}, so only the terms with i=ji=j survive:

⟨u,v⟩=∑k=1nakbk‾.\langle u, v \rangle = \sum_{k=1}^n a_k \overline{b_k}.

This is exactly the claimed componentwise formula. Note this used only orthonormality of the basis and the sesquilinearity axioms already assumed for the inner product on VV; no circularity is involved since we never assumed the conclusion of the theorem to derive it.

Proof of (c). Apply part (b) with u=vu = v. Then ak=bk=⟨v,ek⟩a_k = b_k = \langle v, e_k \rangle for each kk, so

∥v∥2=⟨v,v⟩=∑k=1nakak‾=∑k=1n∣ak∣2=∑k=1n∣⟨v,ek⟩∣2,\|v\|^2 = \langle v, v \rangle = \sum_{k=1}^n a_k \overline{a_k} = \sum_{k=1}^n |a_k|^2 = \sum_{k=1}^n |\langle v, e_k \rangle|^2,

where we used that aa‾=∣a∣2a\overline a = |a|^2 for every scalar a∈Fa \in \mathbb{F} (a basic property of the modulus, true in both R\mathbb R and C\mathbb C). This proves (c). ■\blacksquare

∎
LemmaUniqueness of coordinates relative to any basis

If (e1,…,en)(e_1,\dots,e_n) is a basis (not necessarily orthonormal) of a vector space VV and v∈Vv \in V, then there is exactly one tuple of scalars (a1,…,an)(a_1,\dots,a_n) with v=∑k=1nakekv = \sum_{k=1}^n a_k e_k.

Proof

Existence is the spanning property of a basis. For uniqueness, suppose v=∑kakek=∑kak′ekv = \sum_k a_k e_k = \sum_k a_k' e_k. Subtracting, ∑k(ak−ak′)ek=0\sum_k (a_k - a_k') e_k = 0. Since (e1,…,en)(e_1,\dots,e_n) is linearly independent, every coefficient in a linear combination summing to 00 must be 00; hence ak=ak′a_k = a_k' for all kk. ■\blacksquare

∎

This lemma was invoked silently in part (a) of the theorem above to guarantee that some unique tuple of coefficients exists before we identified it with the Fourier coefficients; it is a basic fact about bases, established independently of orthonormality and inner products, so its use is not circular.

Worked examples

Example 1 — coordinates and inner product in R3\mathbb{R}^3 with a genuinely non-standard orthonormal basis. Let V=R3V = \mathbb{R}^3 with the standard dot product. Consider the three vectors

e1=(23,23,13),e2=(13,−23,23),e3=(23,−13,−23).e_1 = \left(\tfrac{2}{3}, \tfrac{2}{3}, \tfrac{1}{3}\right), \qquad e_2 = \left(\tfrac{1}{3}, -\tfrac{2}{3}, \tfrac{2}{3}\right), \qquad e_3 = \left(\tfrac{2}{3}, -\tfrac{1}{3}, -\tfrac{2}{3}\right).

Step 1: verify orthonormality. Compute norms:

∥e1∥2=49+49+19=1,∥e2∥2=19+49+49=1,∥e3∥2=49+19+49=1.\|e_1\|^2 = \tfrac{4}{9}+\tfrac{4}{9}+\tfrac{1}{9} = 1, \quad \|e_2\|^2 = \tfrac{1}{9}+\tfrac{4}{9}+\tfrac{4}{9} = 1, \quad \|e_3\|^2 = \tfrac{4}{9}+\tfrac{1}{9}+\tfrac{4}{9} = 1.

Compute pairwise inner products:

⟨e1,e2⟩=29−49+29=0,⟨e1,e3⟩=49−29−29=0,⟨e2,e3⟩=29+29−49=0.\langle e_1,e_2\rangle = \tfrac{2}{9} - \tfrac{4}{9} + \tfrac{2}{9} = 0, \qquad \langle e_1,e_3\rangle = \tfrac{4}{9} - \tfrac{2}{9} - \tfrac{2}{9} = 0, \qquad \langle e_2,e_3\rangle = \tfrac{2}{9} + \tfrac{2}{9} - \tfrac{4}{9} = 0.

So (e1,e2,e3)(e_1,e_2,e_3) is orthonormal, and since it consists of 3=dim⁡R33 = \dim \mathbb{R}^3 orthonormal (hence linearly independent — an orthonormal list is automatically linearly independent, since applying ⟨⋅,ej⟩\langle \cdot, e_j\rangle to a dependence relation ∑ciei=0\sum c_ie_i=0 forces cj=0c_j=0 for each jj) vectors, it is a basis, hence an orthonormal basis. Step 2: find the coordinates of v=(1,2,3)v = (1, 2, 3). By the Fourier coefficient formula,

a1=⟨v,e1⟩=23(1)+23(2)+13(3)=2+4+33=3,a_1 = \langle v, e_1 \rangle = \tfrac{2}{3}(1) + \tfrac{2}{3}(2) + \tfrac{1}{3}(3) = \tfrac{2+4+3}{3} = 3,
a2=⟨v,e2⟩=13(1)−23(2)+23(3)=1−4+63=1,a_2 = \langle v, e_2 \rangle = \tfrac{1}{3}(1) - \tfrac{2}{3}(2) + \tfrac{2}{3}(3) = \tfrac{1 - 4 + 6}{3} = 1,
a3=⟨v,e3⟩=23(1)−13(2)−23(3)=2−2−63=−2.a_3 = \langle v, e_3 \rangle = \tfrac{2}{3}(1) - \tfrac{1}{3}(2) - \tfrac{2}{3}(3) = \tfrac{2 - 2 - 6}{3} = -2.

Check: 3e1+1e2−2e3=(2,2,1)+(13,−23,23)+(−43,23,43)=(2+13−43, 2−23+23, 1+23+43)=(1,2,3)=v.3e_1 + 1e_2 - 2e_3 = (2,2,1) + (\tfrac13,-\tfrac23,\tfrac23) + (-\tfrac43,\tfrac23,\tfrac43) = (2+\tfrac13-\tfrac43,\ 2-\tfrac23+\tfrac23,\ 1+\tfrac23+\tfrac43) = (1,2,3) = v. ✓. No linear system was solved: each coordinate came from one dot product. Step 3: compute ∥v∥2\|v\|^2 two ways. Directly, ∥v∥2=1+4+9=14\|v\|^2 = 1+4+9 = 14. Via part (c) of the theorem, ∥v∥2=a12+a22+a32=9+1+4=14\|v\|^2 = a_1^2+a_2^2+a_3^2 = 9+1+4=14. They agree, confirming the componentwise norm formula. Step 4: an inner product computation. Let w=(0,−1,2)w = (0,-1,2). Its coordinates are

b1=⟨w,e1⟩=0−23+23=0,b2=⟨w,e2⟩=0+23+43=2,b3=⟨w,e3⟩=0+13−43=−1.b_1 = \langle w,e_1\rangle = 0 - \tfrac23 + \tfrac23 = 0,\quad b_2 = \langle w,e_2\rangle = 0+\tfrac23+\tfrac43 = 2,\quad b_3 = \langle w,e_3\rangle = 0+\tfrac13-\tfrac43 = -1.

By part (b), ⟨v,w⟩=a1b1+a2b2+a3b3=3(0)+1(2)+(−2)(−1)=0+2+2=4\langle v,w\rangle = a_1b_1+a_2b_2+a_3b_3 = 3(0)+1(2)+(-2)(-1) = 0+2+2 = 4. Directly, ⟨v,w⟩=1(0)+2(−1)+3(2)=−2+6=4\langle v,w\rangle = 1(0)+2(-1)+3(2) = -2+6=4. They match.

Example 2 — a complex orthonormal basis and Fourier coefficients as complex numbers. Let V=C2V = \mathbb{C}^2 with the standard Hermitian inner product ⟨u,v⟩=u1v1‾+u2v2‾\langle u, v\rangle = u_1\overline{v_1} + u_2\overline{v_2}. Take the orthonormal basis

e1=12(1,i),e2=12(i,1).e_1 = \frac{1}{\sqrt2}(1, i), \qquad e_2 = \frac{1}{\sqrt2}(i, 1).

Step 1: verify orthonormality.

∥e1∥2=12(∣1∣2+∣i∣2)=12(1+1)=1,∥e2∥2=12(∣i∣2+∣1∣2)=1.\|e_1\|^2 = \tfrac12(|1|^2+|i|^2) = \tfrac12(1+1)=1,\qquad \|e_2\|^2 = \tfrac12(|i|^2+|1|^2)=1.
⟨e1,e2⟩=12(1⋅i‾+i⋅1‾)=12(1⋅(−i)+i⋅1)=12(−i+i)=0.\langle e_1,e_2\rangle = \tfrac12\big(1\cdot\overline{i} + i\cdot\overline{1}\big) = \tfrac12(1\cdot(-i) + i\cdot 1) = \tfrac12(-i+i) = 0.

So (e1,e2)(e_1,e_2) is orthonormal in C2\mathbb{C}^2, hence (being 2=dim⁡CC22=\dim_{\mathbb C}\mathbb C^2 linearly independent vectors) an orthonormal basis. Step 2: find coordinates of v=(2+i, 1−i)v = (2+i,\ 1-i).

a1=⟨v,e1⟩=12[(2+i)1‾+(1−i)i‾]=12[(2+i)+(1−i)(−i)].a_1 = \langle v, e_1\rangle = \frac{1}{\sqrt2}\Big[(2+i)\overline{1} + (1-i)\overline{i}\Big] = \frac{1}{\sqrt2}\Big[(2+i) + (1-i)(-i)\Big].

Compute (1−i)(−i)=−i+i2⋅(−1)⋅(−1)(1-i)(-i) = -i + i^2\cdot(-1)\cdot(-1); more carefully, (1−i)(−i)=−i−(−i)(i)⋅(−1)(1-i)(-i) = -i - (-i)(i)\cdot(-1) — let's expand directly: (1−i)(−i)=−i⋅1−i⋅(−i)⋅(−1)(1-i)(-i) = -i \cdot 1 - i\cdot(-i)\cdot(-1). Simplify cleanly: (1−i)(−i)=−i+i2=−i−1(1-i)(-i) = -i + i^2 = -i -1 (using i⋅i=i2=−1i\cdot i = i^2=-1, and (−i)(−i)(-i)(-i) is not needed — recompute term by term: (1)(−i)+(−i)(−i)=−i+i2=−i−1(1)(-i) + (-i)(-i) = -i + i^2 = -i -1). So

a1=12[(2+i)+(−i−1)]=12(1)=12.a_1 = \frac{1}{\sqrt2}\big[(2+i) + (-i-1)\big] = \frac{1}{\sqrt2}(1) = \frac{1}{\sqrt2}.
a2=⟨v,e2⟩=12[(2+i)i‾+(1−i)1‾]=12[(2+i)(−i)+(1−i)].a_2 = \langle v, e_2 \rangle = \frac{1}{\sqrt2}\Big[(2+i)\overline{i} + (1-i)\overline{1}\Big] = \frac{1}{\sqrt2}\Big[(2+i)(-i) + (1-i)\Big].

Compute (2+i)(−i)=−2i−i2=−2i+1(2+i)(-i) = -2i - i^2 = -2i+1. So

a2=12[(1−2i)+(1−i)]=12(2−3i).a_2 = \frac{1}{\sqrt2}\big[(1-2i) + (1-i)\big] = \frac{1}{\sqrt2}(2 - 3i).

Check: a1e1+a2e2=12⋅12(1,i)+2−3i2⋅12(i,1)=12(1,i)+2−3i2(i,1)a_1 e_1 + a_2 e_2 = \frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}(1,i) + \frac{2-3i}{\sqrt2}\cdot\frac{1}{\sqrt2}(i,1) = \frac12(1,i) + \frac{2-3i}{2}(i,1). First component: 12+(2−3i)i2=12+2i−3i22=12+2i+32=1+3+2i2=4+2i2=2+i.\frac12 + \frac{(2-3i)i}{2} = \frac12 + \frac{2i - 3i^2}{2} = \frac12 + \frac{2i+3}{2} = \frac{1 + 3 + 2i}{2} = \frac{4+2i}{2} = 2+i. ✓ Second component: i2+2−3i2=i+2−3i2=2−2i2=1−i.\frac{i}{2} + \frac{2-3i}{2} = \frac{i + 2 - 3i}{2} = \frac{2-2i}{2} = 1-i. ✓ Both components recover v=(2+i,1−i)v = (2+i, 1-i) exactly, confirming the coordinate formula in the complex, non-real case. Step 3: norm check. Directly ∥v∥2=∣2+i∣2+∣1−i∣2=(4+1)+(1+1)=5+2=7\|v\|^2 = |2+i|^2 + |1-i|^2 = (4+1)+(1+1) = 5+2=7. Via coordinates: ∣a1∣2=12|a_1|^2 = \tfrac12, and ∣a2∣2=∣2−3i∣22=4+92=132|a_2|^2 = \frac{|2-3i|^2}{2} = \frac{4+9}{2} = \frac{13}{2}. Sum: 12+132=142=7\tfrac12 + \tfrac{13}{2} = \tfrac{14}{2}=7. ✓ This illustrates that in the complex case the correct formula uses ∣⟨v,ek⟩∣2|\langle v,e_k\rangle|^2, not ⟨v,ek⟩2\langle v,e_k\rangle^2 — an easy place for sign/conjugation errors to creep in.

CounterexampleWhy orthonormality (not just linear independence, or just normalization) is essential

Drop orthogonality but keep the vectors a basis and even keep them unit vectors — that is, weaken the hypothesis "(e1,…,en)(e_1,\dots,e_n) orthonormal" to merely "(e1,…,en)(e_1,\dots,e_n) a basis with ∥ek∥=1\|e_k\|=1 for each kk." The coordinate formula then fails. Work in V=R2V=\mathbb{R}^2 with the dot product. Let

e1=(1,0),e2=(12,32).e_1 = (1,0), \qquad e_2 = \left(\tfrac12, \tfrac{\sqrt3}{2}\right).

Both have norm 11: ∥e1∥=1\|e_1\|=1 trivially, and ∥e2∥2=14+34=1\|e_2\|^2 = \tfrac14+\tfrac34=1. They are linearly independent (not parallel), so (e1,e2)(e_1,e_2) is a basis of R2\mathbb R^2. But

⟨e1,e2⟩=1⋅12+0⋅32=12≠0,\langle e_1, e_2\rangle = 1\cdot\tfrac12 + 0\cdot\tfrac{\sqrt3}{2} = \tfrac12 \ne 0,

so the list is not orthogonal, hence not orthonormal. Take v=(1,1)v = (1,1). Its true coordinates (c1,c2)(c_1,c_2) relative to (e1,e2)(e_1,e_2) solve c1e1+c2e2=vc_1 e_1 + c_2 e_2 = v:

c1+12c2=1,32c2=1 ⟹ c2=23,c1=1−13.c_1 + \tfrac12 c_2 = 1, \qquad \tfrac{\sqrt3}{2}c_2 = 1 \ \Longrightarrow\ c_2 = \tfrac{2}{\sqrt3}, \quad c_1 = 1 - \tfrac{1}{\sqrt3}.

Numerically c1≈1−0.577=0.423c_1 \approx 1 - 0.577 = 0.423 and c2≈1.155c_2 \approx 1.155. Now compute the would-be Fourier coefficients, i.e. ⟨v,ek⟩\langle v, e_k\rangle:

⟨v,e1⟩=1⋅1+1⋅0=1,⟨v,e2⟩=1⋅12+1⋅32=12+32≈1.366.\langle v, e_1\rangle = 1\cdot 1 + 1\cdot 0 = 1, \qquad \langle v,e_2\rangle = 1\cdot\tfrac12 + 1\cdot\tfrac{\sqrt3}{2} = \tfrac12+\tfrac{\sqrt3}{2} \approx 1.366.

These numbers, 11 and ≈1.366\approx 1.366, are not equal to the true coordinates ≈0.423\approx 0.423 and ≈1.155\approx 1.155. Indeed

1⋅e1+1.366⋅e2≈(1,0)+1.366(0.5,0.866)≈(1.683, 1.183)≠(1,1)=v.1\cdot e_1 + 1.366\cdot e_2 \approx (1,0) + 1.366(0.5, 0.866) \approx (1.683,\ 1.183) \ne (1,1) = v.

So plugging ⟨v,ek⟩\langle v,e_k\rangle into ∑k⟨v,ek⟩ek\sum_k \langle v,e_k\rangle e_k gives the wrong vector entirely when the basis fails to be orthogonal, even though every vector still has unit length. This shows that unit length alone (without pairwise orthogonality) is not enough for the coordinate formula: the cross terms ⟨ei,ej⟩\langle e_i,e_j\rangle for i≠ji\ne j, which the proof of part (a) relied on vanishing, do not vanish here, and they corrupt the computation.

Scope, edge cases & common traps

SituationWhat happens
VV finite-dimensional, real inner product spaceAll three formulas hold exactly as stated, with real Fourier coefficients.
VV finite-dimensional, complex inner product spaceFormulas hold verbatim, but coefficients are complex; the norm formula needs ∣⋅∣2\lvert\cdot\rvert^2, not squares without absolute value.
(e1,…,ek)(e_1,\dots,e_k) orthonormal but k<dim⁡Vk < \dim V (not a basis)The coordinate formula (a) fails as stated; instead ∑j⟨v,ej⟩ej\sum_j \langle v,e_j\rangle e_j is the orthogonal projection of vv onto span⁡(e1,…,ek)\operatorname{span}(e_1,\dots,e_k), and only equals vv when vv already lies in that span. This is the seed of the Bessel and Parseval Identities.
Basis orthogonal but not normalized (∥ek∥=ck≠1\|e_k\|=c_k\ne1)Must rescale: correct coefficient is ⟨v,ek⟩/∥ek∥2\langle v,e_k\rangle / \|e_k\|^2, not ⟨v,ek⟩\langle v,e_k\rangle directly.
Basis normalized but not orthogonalCoordinate formula fails, as shown in the counterexample above; genuine linear-system solving (or Gram-Schmidt) is required.
Infinite-dimensional inner product space, orthonormal sequenceSums become infinite series; convergence and completeness issues arise — treated properly under Parseval's identity and Hilbert space theory, beyond this article's finite-dimensional scope.

The mathematician behind it

Origins · 1 minJoseph Fourier and how this was discovered
JF
Joseph Fourier
1768–1830 · French mathematician and physicist, Napoleonic-era scientific administrator

Jean-Baptiste Joseph Fourier is best known for his 1822 treatise Théorie analytique de la chaleur ("The Analytical Theory of Heat"), in which he claimed that essentially arbitrary periodic functions could be expanded as infinite sums of sines and cosines. Fourier's original coefficient formulas — obtained by integrating a function against sine and cosine functions — are precisely the infinite-dimensional, function-space analogue of the coordinate formula proved above, with the orthonormal basis (ek)(e_k) replaced by the orthogonal trigonometric system and the finite sum replaced by an infinite series. Though Fourier's work predates the modern abstract definition of an inner product space by nearly a century, the structural insight — that an orthogonal (or orthonormal) system of "basis functions" lets you read off expansion coefficients by a single integral (an inner product) rather than by solving simultaneously for all coefficients at once — is exactly the content of the theorem proved here, transplanted from function spaces to general finite-dimensional inner product spaces.

History
- 1768 — Joseph Fourier born in Auxerre, France.
- 1807–1822 — Fourier develops his theory of heat conduction and trigonometric series expansions, submitting a memoir to the French Academy in 1807 and publishing the full theory in 1822.
- late 19th century — Mathematicians including Hilbert and Schmidt formalize inner product spaces and orthonormal systems abstractly, generalizing Fourier's coefficient formula beyond trigonometric functions.
- 1907–1910s — Development of Hilbert space theory (Hilbert, Riesz, Fischer) rigorously extends orthonormal expansions to infinite dimensions, yielding the Parseval and Riesz–Fischer theorems.
- 20th century — The finite-dimensional coordinate formula becomes a standard early topic in linear algebra curricula, serving as the bridge between abstract inner product spaces and concrete Euclidean coordinate computations.
Does the Fourier coefficient formula require VV to be finite-dimensional?

The theorem as proved here is stated and proved for finite-dimensional VV, where "basis" means a finite linearly independent spanning list and the sum ∑k=1n⟨v,ek⟩ek\sum_{k=1}^n \langle v,e_k\rangle e_k is a genuine finite sum, requiring no analysis. In infinite-dimensional inner product spaces (pre-Hilbert or Hilbert spaces), an analogous formula holds for orthonormal sequences, but the sum becomes an infinite series, and one must address convergence of that series and whether it actually equals vv (completeness of the orthonormal system). That extension is the subject of the Bessel and Parseval identities, which build on exactly this finite-dimensional result but require additional machinery from analysis.

Why does the formula use ⟨v,ek⟩\langle v, e_k\rangle and not ⟨ek,v⟩\langle e_k, v\rangle in the complex case?

In a complex inner product space, the two quantities differ by conjugation: ⟨ek,v⟩=⟨v,ek⟩‾\langle e_k, v\rangle = \overline{\langle v, e_k\rangle}. The convention adopted here — linear in the first argument, conjugate-linear in the second — makes ⟨v,ek⟩\langle v, e_k \rangle the correct coefficient of eke_k in the expansion of vv, as the proof of part (a) shows directly. If your textbook uses the opposite convention (linear in the second argument), the coefficient formula instead reads ⟨ek,v⟩\langle e_k, v \rangle; the mathematics is identical, only the convention for which slot is linear differs, so always check which convention is in force before applying the formula.

Is every basis of a finite-dimensional inner product space "close to" orthonormal?

No, not automatically — as the counterexample demonstrates, a basis of unit vectors can still fail to be orthonormal, and a basis with all pairwise-orthogonal vectors can fail to be normalized. However, every finite-dimensional inner product space does possess an orthonormal basis, and moreover any orthonormal list can be extended to one; this existence result is precisely what the Gram-Schmidt Process establishes constructively, by systematically subtracting off projections to enforce orthogonality and then rescaling to enforce normalization.

How does this relate to matrices being "orthogonal" or "unitary"?

If (e1,…,en)(e_1,\dots,e_n) is an orthonormal basis of Fn\mathbb{F}^n (with the standard inner product) and QQ is the matrix whose columns are e1,…,ene_1,\dots,e_n, then part (b) of the theorem, applied to the standard basis vectors, is exactly the statement that Q∗Q=IQ^*Q = I (where Q∗Q^* is the conjugate transpose), i.e. QQ is unitary (or orthogonal if F=R\mathbb F = \mathbb R). The coordinate formula v=∑k⟨v,ek⟩ekv = \sum_k \langle v,e_k\rangle e_k is then equivalent to the matrix identity v=Q(Q∗v)v = Q(Q^*v), i.e. Q−1=Q∗Q^{-1} = Q^*: recovering coordinates by inner products is the same fact as an orthogonal/unitary matrix having its transpose (conjugate transpose) as its inverse.

What goes wrong if I try to use this formula with an orthogonal, but not orthonormal, basis?

If (f1,…,fn)(f_1,\dots,f_n) is merely orthogonal (pairwise inner products zero) but ∥fk∥≠1\|f_k\| \ne 1 for some kk, the correct coefficient of fkf_k in the expansion of vv is ⟨v,fk⟩/∥fk∥2\langle v, f_k\rangle / \|f_k\|^2, not ⟨v,fk⟩\langle v, f_k\rangle alone. This follows immediately by rerunning the proof of part (a): expanding v=∑akfkv = \sum a_k f_k and taking ⟨⋅,fj⟩\langle \cdot, f_j\rangle gives ⟨v,fj⟩=aj⟨fj,fj⟩=aj∥fj∥2\langle v,f_j\rangle = a_j \langle f_j,f_j\rangle = a_j\|f_j\|^2, so aj=⟨v,fj⟩/∥fj∥2a_j = \langle v,f_j\rangle/\|f_j\|^2. Forgetting to divide by ∥fk∥2\|f_k\|^2 is a common bookkeeping error when working with orthogonal (rather than orthonormal) eigenbases, such as Legendre or Hermite polynomial bases before normalization.

Does the componentwise inner product formula assume uu and vv are expressed in the same orthonormal basis?

Yes, essentially — the formula ⟨u,v⟩=∑kakbk‾\langle u,v\rangle = \sum_k a_k\overline{b_k} requires that ak=⟨u,ek⟩a_k = \langle u,e_k\rangle and bk=⟨v,ek⟩b_k = \langle v,e_k\rangle come from the same list (e1,…,en)(e_1,\dots,e_n). If uu's coordinates were computed relative to one orthonormal basis and vv's relative to a different orthonormal basis, the raw dot product of the two coordinate tuples generally does not equal ⟨u,v⟩\langle u,v\rangle; one must first convert both to a common basis (via a change-of-basis, i.e. unitary/orthogonal, matrix) before the componentwise dot-product shortcut applies.

Before this

Where this leads