Orthonormal Basis¶

Definition¶

Let $V$ be a real vector space with an inner product, written $\boldsymbol{x}\cdot\boldsymbol{y}$ (in $\mathbb{R}^n$ : the dot product). Vectors $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_k\in V$ are orthonormal (form an orthonormal set) if they are pairwise orthogonal and each has norm $1$ :

\boldsymbol{q}_i\cdot\boldsymbol{q}_j=\begin{cases}1,&i=j,\\0,&i\neq j.\end{cases}

An orthonormal basis of $V$ is an orthonormal set that is also a basis of $V$ .

Interpretation¶

An orthonormal basis is a coordinate system with perpendicular axes and unit length along each axis, like the standard basis $\boldsymbol{e}_1,\ldots,\boldsymbol{e}_n$ of $\mathbb{R}^n$ but possibly rotated or reflected. Its advantage is that coordinates no longer require solving a system: the $i$ -th coordinate of $\boldsymbol{x}$ is simply $\boldsymbol{x}\cdot\boldsymbol{q}_i$ $q_{i}$ , the signed length of the orthogonal projection of $\boldsymbol{x}$ onto the line through $\boldsymbol{q}_i$ .

Plane with the orthonormal basis q1, q2 at 45 degrees inside the unit circle and x = (3,1); perpendiculars from x meet the two basis lines at 2√2 q1 and √2 q2.

The figure uses the first example below: $\boldsymbol{q}_1=\tfrac{1}{\sqrt{2}}(1,1)^{\top}$ (blue) and $\boldsymbol{q}_2=\tfrac{1}{\sqrt{2}}(1,-1)^{\top}$ (red) end on the unit circle and meet at a right angle. Dropping perpendiculars from $\boldsymbol{x}=(3,1)^{\top}$ onto the two lines gives the feet $2\sqrt{2}\,\boldsymbol{q}_1=(2,2)^{\top}$ and $\sqrt{2}\,\boldsymbol{q}_2=(1,-1)^{\top}$ $(1, - 1)^{⊤}$ , so the coordinates are read off as the inner products $\boldsymbol{x}\cdot\boldsymbol{q}_1=2\sqrt{2}$ and $\boldsymbol{x}\cdot\boldsymbol{q}_2=\sqrt{2}$ $2$ , and the origin, the two feet and $\boldsymbol{x}$ are the corners of a rectangle. No system of equations is solved.

The two conditions do different jobs. Orthogonality alone gives an orthogonal set, whose coordinates are $(\boldsymbol{x}\cdot\boldsymbol{v}_i)/(\boldsymbol{v}_i\cdot\boldsymbol{v}_i)$ $v_{i})$ ; dividing each $\boldsymbol{v}_i$ by $\|\boldsymbol{v}_i\|$ (normalizing) removes the denominators. Unit length alone is not enough: $(1,0)^{\top}$ and $\tfrac{1}{\sqrt{2}}(1,1)^{\top}$ are unit vectors but not orthogonal.

For the orthonormal pair $\boldsymbol{q}_1=\tfrac{1}{\sqrt{2}}(1,1)^{\top}$ $\frac{1}{2} (1, 1)^{⊤}$ , $\boldsymbol{q}_2=\tfrac{1}{\sqrt{2}}(1,-1)^{\top}$ and $\boldsymbol{x}=(3,1)^{\top}$ $(3, 1)^{⊤}$ , the inner products $2\sqrt{2}$ and $\sqrt{2}$ are the coordinates, and $(\boldsymbol{x}\cdot\boldsymbol{q}_1)\boldsymbol{q}_1+(\boldsymbol{x}\cdot\boldsymbol{q}_2)\boldsymbol{q}_2$ lands exactly on $\boldsymbol{x}$ . Switch to an orthogonal pair that is not unit, or to two unit vectors that are not orthogonal: the same recipe misses $\boldsymbol{x}$ , and $Q^{\top}Q$ shows which condition fails.

In matrix form, put $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_k\in\mathbb{R}^n$ as the columns of an $n\times k$ matrix $Q$ . The $(i,j)$ entry of $Q^{\top}Q$ is $\boldsymbol{q}_i\cdot\boldsymbol{q}_j$ $q_{j}$ , so the columns are orthonormal exactly when $Q^{\top}Q=I_k$ $I_{k}$ . When $k=n$ $n$ , $Q$ is an orthogonal matrix and is the transformation matrix of the basis in the sense of change of basis; its inverse is simply $Q^{\top}$ .

Key Results¶

An orthonormal set is linearly independent: taking the inner product of $\displaystyle \sum_i c_i\boldsymbol{q}_i=\boldsymbol{0}$ with $\boldsymbol{q}_j$ gives $c_j=0$ $0$ . The same argument works for any orthogonal set of nonzero vectors. Hence an orthonormal set in an $n$ -dimensional space has at most $n$ vectors, and $n$ orthonormal vectors form an orthonormal basis.
Expansion: if $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_n$ is an orthonormal basis of $V$ , every $\boldsymbol{x}\in V$ satisfies $\displaystyle \boldsymbol{x}=\sum_{i=1}^n(\boldsymbol{x}\cdot\boldsymbol{q}_i)\,\boldsymbol{q}_i$ $q_{i}) q_{i}$ ; in $\mathbb{R}^n$ the coordinate vector is $\boldsymbol{y}=Q^{\top}\boldsymbol{x}$ $Q^{⊤} x$ .
Lengths and inner products can be computed in coordinates: $\displaystyle \boldsymbol{x}\cdot\boldsymbol{z}=\sum_i(\boldsymbol{x}\cdot\boldsymbol{q}_i)(\boldsymbol{z}\cdot\boldsymbol{q}_i)$ and $\displaystyle \|\boldsymbol{x}\|^2=\sum_i(\boldsymbol{x}\cdot\boldsymbol{q}_i)^2$ (Parseval's identity). With respect to a basis that is not orthonormal, neither formula holds.
Existence: every finite-dimensional inner product space has an orthonormal basis, and every orthonormal set extends to one; the Gram-Schmidt algorithm constructs it from any basis.
If $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_k$ is an orthonormal basis of a subspace $W$ of $\mathbb{R}^n$ , then $\displaystyle \operatorname{proj}_W\boldsymbol{x}=\sum_{i=1}^k(\boldsymbol{x}\cdot\boldsymbol{q}_i)\boldsymbol{q}_i=QQ^{\top}\boldsymbol{x}$ $Q Q^{⊤} x$ ; for $k<n$ $n$ , $QQ^{\top}\neq I_n$ even though $Q^{\top}Q=I_k$ $I_{k}$ .
If $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_k$ is extended to an orthonormal basis $\boldsymbol{q}_1,\ldots,\boldsymbol{q}_n$ of $\mathbb{R}^n$ , the added vectors $\boldsymbol{q}_{k+1},\ldots,\boldsymbol{q}_n$ form an orthonormal basis of the orthogonal complement $W^{\perp}$ , so $\dim W+\dim W^{\perp}=n$ $n$ .

Examples¶

Coordinates without inverting. In $\mathbb{R}^2$ , $\boldsymbol{q}_1=\tfrac{1}{\sqrt{2}}(1,1)^{\top}$ and $\boldsymbol{q}_2=\tfrac{1}{\sqrt{2}}(1,-1)^{\top}$ are orthonormal: $\boldsymbol{q}_1\cdot\boldsymbol{q}_2=\tfrac{1}{2}(1-1)=0$ and $\boldsymbol{q}_1\cdot\boldsymbol{q}_1=\boldsymbol{q}_2\cdot\boldsymbol{q}_2=1$ $1$ . For $\boldsymbol{x}=(3,1)^{\top}$ $(3, 1)^{⊤}$ ,

\begin{aligned} \boldsymbol{x}\cdot\boldsymbol{q}_1&=\tfrac{4}{\sqrt{2}}=2\sqrt{2},\\ \boldsymbol{x}\cdot\boldsymbol{q}_2&=\tfrac{2}{\sqrt{2}}=\sqrt{2}, \end{aligned}

and indeed $2\sqrt{2}\,\boldsymbol{q}_1+\sqrt{2}\,\boldsymbol{q}_2=(2,2)^{\top}+(1,-1)^{\top}=(3,1)^{\top}$ $(3, 1)^{⊤}$ . The length check gives $(2\sqrt{2})^2+(\sqrt{2})^2=10=3^2+1^2$ $1^{2}$ .

Normalizing an orthogonal set. In $\mathbb{R}^3$ the vectors $(1,1,0)^{\top}$ , $(1,-1,0)^{\top}$ , $(0,0,2)^{\top}$ are pairwise orthogonal with norms $\sqrt{2}$ , $\sqrt{2}$ , $2$ . Dividing by the norms gives the orthonormal basis

\begin{aligned} \boldsymbol{q}_1&=\tfrac{1}{\sqrt{2}}(1,1,0)^{\top},\\ \boldsymbol{q}_2&=\tfrac{1}{\sqrt{2}}(1,-1,0)^{\top},\\ \boldsymbol{q}_3&=(0,0,1)^{\top}. \end{aligned}

Projecting with an orthonormal basis. For $W=\operatorname{span}\{\boldsymbol{q}_1,\boldsymbol{q}_2\}$ from the previous example (the plane $x_3=0$ $0$ ) and $\boldsymbol{b}=(2,0,5)^{\top}$ $(2, 0, 5)^{⊤}$ : $\boldsymbol{b}\cdot\boldsymbol{q}_1=\sqrt{2}$ and $\boldsymbol{b}\cdot\boldsymbol{q}_2=\sqrt{2}$ $2$ , so $\operatorname{proj}_W\boldsymbol{b}=(1,1,0)^{\top}+(1,-1,0)^{\top}=(2,0,0)^{\top}$ $(2, 0, 0)^{⊤}$ , as expected for the plane $x_3=0$ $0$ .

A tilted plane and its complement. The vectors $\boldsymbol{q}_1=\tfrac{1}{3}(1,2,2)^{\top}$ and $\boldsymbol{q}_2=\tfrac{1}{3}(2,1,-2)^{\top}$ are orthonormal: $\boldsymbol{q}_1\cdot\boldsymbol{q}_2=\tfrac{1}{9}(2+2-4)=0$ and both have squared length $\tfrac{1}{9}(1+4+4)=1$ $1$ . They span a plane $W$ through the origin. With $Q=[\boldsymbol{q}_1\ \boldsymbol{q}_2]$ $[q_{1} q_{2}]$ , $Q^{\top}Q=I_2$ $I_{2}$ , but

QQ^{\top}=\frac{1}{9}\begin{pmatrix}5&4&-2\\4&5&2\\-2&2&8\end{pmatrix}\neq I_3.

For $\boldsymbol{x}=(0,0,9)^{\top}$ the coordinates are $\boldsymbol{x}\cdot\boldsymbol{q}_1=6$ and $\boldsymbol{x}\cdot\boldsymbol{q}_2=-6$ $- 6$ , so $\operatorname{proj}_W\boldsymbol{x}=6\boldsymbol{q}_1-6\boldsymbol{q}_2=(-2,2,8)^{\top}$ $(- 2, 2, 8)^{⊤}$ . The remainder $(2,-2,1)^{\top}$ is orthogonal to both $\boldsymbol{q}_i$ , and normalizing it gives $\boldsymbol{q}_3=\tfrac{1}{3}(2,-2,1)^{\top}$ $\frac{1}{3} (2, - 2, 1)^{⊤}$ , which completes an orthonormal basis of $\mathbb{R}^3$ and spans $W^{\perp}$ .

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 6 (placement provisional; not yet checked against the slides)

Aliases: orthonormal set, orthonormal vectors, orthonormal list, orthonormal system, ONB, orthonormal Keywords: q_i · q_j = 0 for i ≠ j, q_i · q_i = 1, Q^T Q = I, x = Σ (x · q_i) q_i, Fourier coefficients, Parseval identity, unit vectors, normalize Symbols: element of, real numbers, square root, sum

Last updated: 17 October 2026 21:53