Coefficients Along an Orthonormal List

The Coefficient Along One Unit Vector

An orthonormal list gives a simple way to identify the scalar multiplying any one of its vectors. The key is to take an inner product with that particular vector: orthogonality makes the other list members contribute zero, while unit length leaves the desired scalar unchanged. Over \(\mathbb{C}\), the order of the inner-product arguments matters, so the formula depends on the convention for which slot is linear.

Throughout, the scalar field is \(\mathbb{R}\) or \(\mathbb{C}\). We use the convention that the inner product is linear in its first slot and conjugate-linear in its second slot. Thus \(\langle a\mathbf{x},\mathbf{y}\rangle=a\langle\mathbf{x},\mathbf{y}\rangle\) and \(\langle\mathbf{x},a\mathbf{y}\rangle=\overline{a}\langle\mathbf{x},\mathbf{y}\rangle\). Over \(\mathbb{R}\), conjugation has no effect. In either field, \(\langle\mathbf{u},\mathbf{u}\rangle=\|\mathbf{u}\|^2\).

For a unit vector \(\mathbf{u}\), the coefficient along \(\mathbf{u}\) is \(\langle\mathbf{x},\mathbf{u}\rangle\). The unit-length condition matters: it is what turns \(\langle\mathbf{u},\mathbf{u}\rangle\) into \(1\). For an orthonormal list, taking the inner product with \(\mathbf{u}_k\) isolates the coefficient of that member, provided the other part of the vector is orthogonal to the list.

Why the Formula Gives the Coefficient

Proof. Set \(\alpha=\langle\mathbf{x},\mathbf{u}\rangle\) and \(\mathbf{w}=\mathbf{x}-\alpha\mathbf{u}\). Since the inner product is linear in its first slot and \(\langle\mathbf{u},\mathbf{u}\rangle=1\),

So this choice of \(\alpha\) makes the remainder orthogonal to \(\mathbf{u}\). To prove uniqueness, suppose also that \(\mathbf{x}-\beta\mathbf{u}\) is orthogonal to \(\mathbf{u}\). Then \(\langle\mathbf{x},\mathbf{u}\rangle-\beta\langle\mathbf{u},\mathbf{u}\rangle=0\), so \(\langle\mathbf{x},\mathbf{u}\rangle-\beta=0\), and hence \(\beta=\langle\mathbf{x},\mathbf{u}\rangle=\alpha\). \(\square\)

This result also identifies the vector component along \(\mathbf{u}\): it is \(\langle\mathbf{x},\mathbf{u}\rangle\mathbf{u}\). The rest, \(\mathbf{x}-\langle\mathbf{x},\mathbf{u}\rangle\mathbf{u}\), is orthogonal to \(\mathbf{u}\). This is a statement about one direction at a time; it does not require that \(\mathbf{u}\) span the whole space.

Isolating a Member of an Orthonormal List

Proof. Fix an index \(k\). Take the inner product of the stated decomposition with \(\mathbf{u}_k\) in the second slot. Linearity in the first slot gives

The residual term is zero because \(\mathbf{r}\) is orthogonal to \(\mathbf{u}_k\). Orthonormality gives \(\langle\mathbf{u}_j,\mathbf{u}_k\rangle=0\) for \(j\ne k\) and \(\langle\mathbf{u}_k,\mathbf{u}_k\rangle=1\). The sum therefore has only one nonzero possible term, namely \(c_k\), so \(\langle\mathbf{x},\mathbf{u}_k\rangle=c_k\). Since \(k\) was arbitrary, this holds for each member of the list. \(\square\)

The orthogonality of the residual is part of the hypothesis. If \(\mathbf{x}\) has additional components outside the span of the list, the theorem still applies when those components together form a residual orthogonal to every list member. It does not claim that every vector is a linear combination of the list. Also, the formula extracts a coefficient only relative to such a decomposition; the theorem does not require that the list be a basis of the whole space.

The Complex Convention and Conjugation

Over \(\mathbb{C}\), the order in the formula is essential. With the convention used here, multiplying the first argument by a scalar pulls that scalar out unchanged. Therefore the coefficient \(\alpha\) in \(\alpha\mathbf{u}\) is recovered by \(\langle\alpha\mathbf{u},\mathbf{u}\rangle=\alpha\), not by its conjugate. Reversing the arguments generally conjugates the result: conjugate symmetry gives \(\langle\mathbf{u},\mathbf{x}\rangle=\overline{\langle\mathbf{x},\mathbf{u}\rangle}\).

Common Mistakes and a Useful Contrast

Reversing the arguments in a complex space. In the preceding example, a tempting calculation is to take \(\langle\mathbf{u},\mathbf{x}\rangle\) as the coefficient. It gives \(-i/\sqrt{2}\), not \(i/\sqrt{2}\). Indeed, multiplying \(\mathbf{u}\) by that mistaken scalar would not produce a remainder orthogonal to \(\mathbf{u}\). The correction is to use \(\langle\mathbf{x},\mathbf{u}\rangle\), consistent with linearity in the first slot, and then check that \(\langle\mathbf{x}-\langle\mathbf{x},\mathbf{u}\rangle\mathbf{u},\mathbf{u}\rangle=0\).

Forgetting that the vector must be unit length. If \(\mathbf{v}\ne\mathbf{0}\) is not a unit vector and \(\mathbf{x}=\alpha\mathbf{v}+\mathbf{w}\) with \(\mathbf{w}\perp\mathbf{v}\), then \(\langle\mathbf{x},\mathbf{v}\rangle=\alpha\|\mathbf{v}\|^2\). Thus the coefficient is \(\langle\mathbf{x},\mathbf{v}\rangle/\|\mathbf{v}\|^2\), not simply \(\langle\mathbf{x},\mathbf{v}\rangle\). For example, with \(\mathbf{v}=(2,0)^{\top}\) and \(\mathbf{x}=(3,1)^{\top}\), the coefficient is \(3/2\): the remainder \(\mathbf{x}-(3/2)\mathbf{v}=(0,1)^{\top}\) is orthogonal to \(\mathbf{v}\). The raw inner product is \(6\), which is not the coefficient because \(\|\mathbf{v}\|^2=4\).

For an orthonormal list, the rule is especially clean: take the inner product of the vector with the chosen list member in the second slot. Unit length removes a norm factor, and orthogonality removes contributions from the other list members when the remaining part is orthogonal to the list. In complex spaces, preserving the argument order ensures the coefficient itself—not its conjugate—is recovered.