Coordinates From Projection Onto An Orthonormal Basis

Projection Coefficients as Coordinates

A projection onto a subspace can be described by its components along an orthonormal basis of that subspace. The key simplification is that each component is read off with one inner product: no system of equations is needed. Those inner products are not merely ingredients in the computation; they are the coordinates of the projected vector in the chosen basis.

Let \(V\) be a finite-dimensional inner product space over \(\mathbb{R}\) or \(\mathbb{C}\), with inner product linear in the first slot. In \(\mathbb{R}^n\), this is the usual dot product. An ordered list \(\mathcal{B}=(\mathbf{q}_1,\dots,\mathbf{q}_k)\) is an orthonormal basis of a subspace \(W\) if it spans \(W\) and satisfies \(\langle\mathbf{q}_i,\mathbf{q}_j\rangle=0\) for \(i\ne j\) and \(\langle\mathbf{q}_i,\mathbf{q}_i\rangle=1\) for each \(i\). The order matters because coordinates are listed in that order.

Proof. Define \(\mathbf{p}\) by the displayed sum. It is in \(W\) because it is a linear combination of the basis vectors of \(W\). For each \(j\), orthonormality gives $$ \langle\mathbf{p},\mathbf{q}_j\rangle =\sum_{i=1}^{k}\langle\mathbf{x},\mathbf{q}_i\rangle\langle\mathbf{q}_i,\mathbf{q}_j\rangle =\langle\mathbf{x},\mathbf{q}_j\rangle. $$ Here the inner product is linear in the first slot, so the coefficient \(\langle\mathbf{x},\mathbf{q}_i\rangle\) factors out as written. Therefore \(\langle\mathbf{x}-\mathbf{p},\mathbf{q}_j\rangle=0\) for every \(j\). Every \(\mathbf{w}\in W\) has the form \(\sum_{j=1}^{k}a_j\mathbf{q}_j\); conjugate-linearity in the second slot gives \(\langle\mathbf{x}-\mathbf{p},\mathbf{w}\rangle=\sum_{j=1}^{k}\overline{a_j}\langle\mathbf{x}-\mathbf{p},\mathbf{q}_j\rangle=0\). Thus the residual is orthogonal to all of \(W\). The coefficients in the defining sum are the coordinates of \(\mathbf{p}\), because \(\mathcal{B}\) is a basis and coordinates are the unique coefficients in a basis expansion. \(\square\)

If \(W=\{\mathbf{0}\}\), the basis is the empty list and the sum is the empty sum, equal to \(\mathbf{0}\). Its coordinate list is empty. The formula therefore covers this case as well.

Why These Coefficients Are the Right Ones

The formula follows from two defining properties. First, the projected vector must lie in \(W\), so it must be expressible using the basis vectors. Second, the difference from the original vector must be orthogonal to \(W\). Taking an inner product with \(\mathbf{q}_j\) isolates the \(j\)th coefficient because \(\mathbf{q}_j\) has inner product \(1\) with itself and \(0\) with every other basis vector.

Proof. Suppose \(\mathbf{p}\) and \(\mathbf{p}'\) both lie in \(W\), and both \(\mathbf{x}-\mathbf{p}\) and \(\mathbf{x}-\mathbf{p}'\) are orthogonal to \(W\). Their difference \(\mathbf{p}-\mathbf{p}'\) lies in \(W\), and it is also orthogonal to \(W\), since it equals \((\mathbf{x}-\mathbf{p}')-(\mathbf{x}-\mathbf{p})\). Hence it is orthogonal to itself. Positive definiteness of the inner product implies \(\langle\mathbf{p}-\mathbf{p}',\mathbf{p}-\mathbf{p}'\rangle=0\) only when \(\mathbf{p}-\mathbf{p}'=\mathbf{0}\). Thus \(\mathbf{p}=\mathbf{p}'\). If \(\mathbf{x}\in W\), its projection is \(\mathbf{x}\), because \(\mathbf{x}-\mathbf{x}=\mathbf{0}\) is orthogonal to \(W\); the coordinate statement then follows from the theorem. \(\square\)

This distinction matters: when \(\mathbf{x}\) is outside \(W\), the coefficient list does not give coordinates for \(\mathbf{x}\) in \(\mathcal{B}\), since \(\mathcal{B}\) is a basis of \(W\), not necessarily of the whole space \(V\). It gives coordinates for the particular vector \(\mathbf{p}\in W\).

Computing and Checking the Projection

What Changes for an Orthogonal List

Normalization is essential to the coefficient formula. For an orthonormal basis, each basis vector has squared norm \(1\), so the coefficient is just the inner product. If the vectors are orthogonal but not unit length, the coefficient along \(\mathbf{u}_i\) must instead be \(\langle\mathbf{x},\mathbf{u}_i\rangle/\langle\mathbf{u}_i,\mathbf{u}_i\rangle\). The denominator corrects for the vector's length. This is the distinction between an orthogonal list and an orthonormal one; the latter is the setting of the theorem above.

Common mistake: using the orthonormal formula on an unnormalized vector. Take \(\mathbf{u}=\begin{bmatrix}2\\0\end{bmatrix}\) and \(\mathbf{x}=\begin{bmatrix}3\\1\end{bmatrix}\). A mistaken calculation uses \(\langle\mathbf{x},\mathbf{u}\rangle=6\) as the coefficient and gives \(6\mathbf{u}=\begin{bmatrix}12\\0\end{bmatrix}\). But the residual \(\begin{bmatrix}-9\\1\end{bmatrix}\) has dot product \(-18\) with \(\mathbf{u}\), so it is not orthogonal to the line. The correct coefficient is \(6/\langle\mathbf{u},\mathbf{u}\rangle=6/4=3/2\), giving the projection \((3/2)\mathbf{u}=\begin{bmatrix}3\\0\end{bmatrix}\). Its residual is \(\begin{bmatrix}0\\1\end{bmatrix}\), whose dot product with \(\mathbf{u}\) is \(0\).

Using the Formula Reliably

For a given orthonormal basis, the calculation has a consistent order: compute one inner product for each basis vector, place those values in the basis order, and use the same values as multipliers in the basis expansion. A direct check catches both arithmetic and conceptual errors. Verify that the reconstructed vector lies in the subspace and that the residual has inner product zero with every basis vector. Since those vectors span the subspace, these checks establish residual orthogonality throughout the subspace.

Over \(\mathbb{C}\), keep the stated inner-product convention in mind. With the inner product linear in the first slot, the coefficient of \(\mathbf{q}_i\) is \(\langle\mathbf{x},\mathbf{q}_i\rangle\). Reversing the slots would conjugate this value; using the wrong slot can therefore give incorrect coefficients. In real spaces the inner product is symmetric, so this distinction does not change the numerical result.