From Orthonormality to Coordinates
A basis describes every vector in a space, but the coefficients in a basis expansion are not always easy to read off. An orthonormal basis has a special advantage: each coefficient can be found by taking one inner product. The reason is that inner products with the basis vectors separate the terms in the expansion, leaving just the matching coefficient.
Let \(V\) be a finite-dimensional inner product space over \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). In the complex case, use the convention that the inner product is linear in its first slot and conjugate-linear in its second. An ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) is orthonormal if \(\langle\mathbf{b}_j,\mathbf{b}_k\rangle=0\) whenever \(j\ne k\), and \(\langle\mathbf{b}_j,\mathbf{b}_j\rangle=1\) for every \(j\). Together these conditions say \(\langle\mathbf{b}_j,\mathbf{b}_k\rangle=1\) when \(j=k\), and \(0\) otherwise.
Because \(\mathcal{B}\) is a basis, every \(\mathbf{x}\in V\) has a unique expansion \(\mathbf{x}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), with scalars \(c_1,\dots,c_n\in\mathbb{F}\). The ordered list of these scalars is the coordinate vector \([\mathbf{x}]_{\mathcal{B}}=(c_1,\dots,c_n)^{\top}\). The result below identifies each \(c_k\) directly from \(\mathbf{x}\).
Proof. Since \(\mathcal{B}\) is a basis, write \(\mathbf{x}=\sum_{j=1}^{n}c_j\mathbf{b}_j\), where the \(c_j\) are its unique coordinates. Fix any index \(k\) with \(1\le k\le n\). Linearity in the first slot gives $$ \langle\mathbf{x},\mathbf{b}_k\rangle = \left\langle\sum_{j=1}^{n}c_j\mathbf{b}_j,\mathbf{b}_k\right\rangle = \sum_{j=1}^{n}c_j\langle\mathbf{b}_j,\mathbf{b}_k\rangle. $$ Orthonormality makes every term in this sum zero except the term with \(j=k\), which is \(c_k\langle\mathbf{b}_k,\mathbf{b}_k\rangle=c_k\). Hence \(c_k=\langle\mathbf{x},\mathbf{b}_k\rangle\). This holds for every \(k\), so these inner products are exactly the entries of \([\mathbf{x}]_{\mathcal{B}}\). Substituting them for the \(c_k\) in the basis expansion gives the stated formula for \(\mathbf{x}\). \(\square\)
The proof depends on both parts of the hypothesis. Orthonormality isolates a coefficient, while the basis property guarantees that there is an expansion of every vector in the first place. If \(V=\{\mathbf{0}\}\), its basis is the empty list; the corresponding expansion is the empty sum, equal to \(\mathbf{0}\). The displayed coordinate formula applies to nonzero finite-dimensional spaces, where \(n\ge1\).
Using the Expansion in a Real Space
Worked Example: A Rotated Orthonormal Basis in \(\mathbb{R}^2\)
Let $$ \mathbf{b}_1=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}, \qquad \mathbf{b}_2=\frac{1}{\sqrt{2}}\begin{bmatrix}-1\\1\end{bmatrix}, \qquad \mathbf{x}=\begin{bmatrix}2\\1\end{bmatrix}. $$ Find \([\mathbf{x}]_{\mathcal{B}}\) and expand \(\mathbf{x}\), where \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\).
First check the orthonormality needed to use the theorem. Each vector has squared norm \(1/2+1/2=1\), and \(\mathbf{b}_1\cdot\mathbf{b}_2=(-1+1)/2=0\). The determinant of the matrix with these vectors as columns is \((1\cdot1-(-1)\cdot1)/2=1\), so they form a basis of \(\mathbb{R}^2\).
In \(\mathbb{R}^2\), the inner product is the dot product. Therefore, $$ \langle\mathbf{x},\mathbf{b}_1\rangle =\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}}, \qquad \langle\mathbf{x},\mathbf{b}_2\rangle =\frac{-2+1}{\sqrt{2}}=-\frac{1}{\sqrt{2}}. $$ These are the coordinates, in the order matching the basis:
To check the expansion, calculate each scaled vector: \(\frac{3}{\sqrt{2}}\mathbf{b}_1=(3/2,3/2)^{\top}\) and \(-\frac{1}{\sqrt{2}}\mathbf{b}_2=(1/2,-1/2)^{\top}\). Their sum is \((3/2+1/2,3/2-1/2)^{\top}=(2,1)^{\top}=\mathbf{x}\), as required.
The coefficients depend on the order of the basis: the first entry belongs to \(\mathbf{b}_1\), and the second belongs to \(\mathbf{b}_2\). The expansion formula also reconstructs the vector itself, not just its coordinate list. Keeping those two objects distinct makes it easier to check both the numerical coefficients and their placement.
The Complex Inner-Product Convention
Over \(\mathbb{C}\), conjugate-linearity in the second slot matters. The theorem uses \(\langle\mathbf{x},\mathbf{b}_k\rangle\), with the basis vector in the second slot. This choice returns the coefficient under the stated convention: in an expansion \(\mathbf{x}=\sum_j c_j\mathbf{b}_j\), the coefficient \(c_j\) stays unchanged when the inner product is expanded in its first slot.
Worked Example: Recovering Complex Coordinates
On \(\mathbb{C}^2\), take \(\langle\mathbf{u},\mathbf{v}\rangle=u_1\overline{v_1}+u_2\overline{v_2}\), which is linear in the first slot. Let $$ \mathbf{b}_1=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\i\end{bmatrix}, \qquad \mathbf{b}_2=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\-i\end{bmatrix}, \qquad \mathbf{x}=\begin{bmatrix}1+i\\2\end{bmatrix}. $$ Find the coordinates and verify the expansion.
The squared norm of each basis vector is \((1+|i|^2)/2=1\). Their inner product is \(\langle\mathbf{b}_1,\mathbf{b}_2\rangle=(1+i\overline{-i})/2=(1+i^2)/2=0\), since \(\overline{-i}=i\). The two columns are independent because they are nonzero and orthogonal; in \(\mathbb{C}^2\), two independent vectors form a basis. Thus \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\) is an orthonormal basis.
For the first coordinate, conjugating the entries of \(\mathbf{b}_1\) gives $$ \langle\mathbf{x},\mathbf{b}_1\rangle =\frac{(1+i)\cdot 1+2\cdot(-i)}{\sqrt{2}} =\frac{1-i}{\sqrt{2}}. $$ For the second coordinate, the conjugate of \(-i\) is \(i\), so $$ \langle\mathbf{x},\mathbf{b}_2\rangle =\frac{(1+i)\cdot 1+2\cdot i}{\sqrt{2}} =\frac{1+3i}{\sqrt{2}}. $$ Hence $$ [\mathbf{x}]_{\mathcal{B}} = \begin{bmatrix} (1-i)/\sqrt{2}\\ (1+3i)/\sqrt{2} \end{bmatrix}, \qquad \mathbf{x}=\frac{1-i}{\sqrt{2}}\mathbf{b}_1+ \frac{1+3i}{\sqrt{2}}\mathbf{b}_2. $$
For a direct check, the first scaled vector is \(\big((1-i)/2,(1+i)/2\big)^{\top}\), because \(i(1-i)=i+1\). The second is \(\big((1+3i)/2,(3-i)/2\big)^{\top}\), because \(-i(1+3i)=3-i\). Adding gives \(\big((2+2i)/2,4/2\big)^{\top}=(1+i,2)^{\top}=\mathbf{x}\).
Coordinates Are Unique and Depend Linearly on the Vector
Uniqueness is built into the definition of a basis: a vector has only one list of coefficients in that basis. The inner-product formula gives those coefficients directly. It also shows that the coordinate vector respects addition and scalar multiplication. This property is useful when a vector is described as a sum or scalar multiple of other vectors.
Proof. For each \(k\), the coordinate formula gives the \(k\)th entry of \([\mathbf{x}+\mathbf{y}]_{\mathcal{B}}\) as \(\langle\mathbf{x}+\mathbf{y},\mathbf{b}_k\rangle =\langle\mathbf{x},\mathbf{b}_k\rangle+\langle\mathbf{y},\mathbf{b}_k\rangle\), by linearity in the first slot. These are the \(k\)th entries of the two coordinate vectors on the right, so the vectors of coordinates are equal. Likewise, the \(k\)th entry of \([a\mathbf{x}]_{\mathcal{B}}\) is \(\langle a\mathbf{x},\mathbf{b}_k\rangle=a\langle\mathbf{x},\mathbf{b}_k\rangle\). This holds for every \(k\), proving the scalar identity, including \(a=0\). \(\square\)
Why an Orthonormal List Must Be a Basis
For any vector and any orthonormal list, the inner products with the list still give coefficients for the components along those listed vectors. But without a basis, the resulting sum need not equal the original vector. The theorem guarantees equality because a basis spans the entire space. The following example makes the distinction explicit.
Worked Example: An Orthonormal List That Does Not Expand Every Vector
In \(\mathbb{R}^3\), take the orthonormal list \((\mathbf{e}_1,\mathbf{e}_2)\), where \(\mathbf{e}_1=(1,0,0)^{\top}\) and \(\mathbf{e}_2=(0,1,0)^{\top}\). Use inner products with these vectors to form a sum for \(\mathbf{x}=(2,-1,3)^{\top}\), then check whether it equals \(\mathbf{x}\).
The coefficients are \(\langle\mathbf{x},\mathbf{e}_1\rangle=2\) and \(\langle\mathbf{x},\mathbf{e}_2\rangle=-1\). The sum constructed from the list is $$ 2\mathbf{e}_1-\mathbf{e}_2 = \begin{bmatrix}2\\0\\0\end{bmatrix} + \begin{bmatrix}0\\-1\\0\end{bmatrix} = \begin{bmatrix}2\\-1\\0\end{bmatrix}. $$ Subtracting this from \(\mathbf{x}\) gives \((0,0,3)^{\top}\), which is nonzero. Thus the sum is not \(\mathbf{x}\). The list is orthonormal but does not span \(\mathbb{R}^3\), so it is not a basis. The residual is orthogonal to both listed vectors: its dot products with \(\mathbf{e}_1\) and \(\mathbf{e}_2\) are both \(0\).
Common Pitfall: Reversing the Inner-Product Slots
In a real inner product space, reversing the slots does not change the inner product, so a reversed-slot error can go unnoticed. In a complex inner product space, \(\langle\mathbf{u},\mathbf{v}\rangle=\overline{\langle\mathbf{v},\mathbf{u}\rangle}\); the order matters. Under the convention used here, the coefficient is \(\langle\mathbf{x},\mathbf{b}_k\rangle\), not \(\langle\mathbf{b}_k,\mathbf{x}\rangle\).
For a concrete wrong calculation, let \(\mathbf{x}=i\mathbf{e}_1\) and \(\mathbf{b}_1=\mathbf{e}_1\) in \(\mathbb{C}^2\) with the inner product above. Reversing the slots gives \(\langle\mathbf{e}_1,\mathbf{x}\rangle=1\cdot\overline{i}=-i\). Treating this as the coefficient would produce \(-i\mathbf{e}_1\), which is not \(\mathbf{x}\). The correct order gives \(\langle\mathbf{x},\mathbf{e}_1\rangle=i\cdot\overline{1}=i\), so the expansion is \(i\mathbf{e}_1=\mathbf{x}\). A useful check is to substitute the proposed coefficients back into the basis expansion.
The central use of an orthonormal basis is therefore precise: orthonormality extracts each coefficient, and the basis property ensures the resulting finite sum recovers every vector. For a basis \(\mathcal{B}\), first take \(\langle\mathbf{x},\mathbf{b}_k\rangle\) for each basis vector, keeping the order consistent in the complex case; then use those values as the coefficients of the same ordered basis vectors in the expansion.
Check Your Understanding
Use the orthonormal coordinate formula, and check any proposed expansion by substitution.
- State the two conditions that make an ordered basis orthonormal.
- In the proof of the coordinate formula, why does only one term remain when \(\mathbf{x}\) is paired with \(\mathbf{b}_k\)?
- For the real basis in the first worked example, what is the second coordinate of \(\mathbf{x}\), and how is it computed?
- Why does \(\langle\mathbf{x},\mathbf{b}_k\rangle\) give the coefficient in the complex convention used here?
- For \(\mathbf{x}=(2,-1,3)^{\top}\), why does the list \((\mathbf{e}_1,\mathbf{e}_2)\) fail to give a basis expansion?
