Existence of Orthonormal Bases

Orthonormal bases underlie the QR decomposition used throughout numerical linear algebra to solve least-squares problems stably; the columns of $Q$ are precisely an orthonormal basis obtained by Gram-Schmidt (or a numerically stabilized variant) applied to the columns of a matrix. In physics, orthonormal bases of eigenstates are the mathematical backbone of quantum mechanics, where physical observables correspond to self-adjoint operators diagonalized in an orthonormal basis. In statistics and machine learning, principal component analysis constructs an orthonormal basis of eigenvectors of a covariance matrix to decorrelate data. In signal processing, orthonormal wavelet and Fourier bases decompose signals into independent frequency or scale components. In cryptography, lattice-based schemes rely on the LLL algorithm, a computational descendant of Gram-Schmidt orthogonalization, to find short, nearly-orthogonal lattice vectors. Across all these settings, the existence theorem is the quiet guarantee that the coordinate system engineers and scientists want to use is always available in principle, before any algorithm is run to actually compute it.

Ask the tutor: How does extending an orthonormal basis of a subspace let us explicitly construct its orthogonal complement?

An orthonormal basis is the linear-algebra analogue of a rigid, perpendicular coordinate grid: each axis has been rescaled to unit length, and no axis leans on any other. If a vector's coordinates with respect to an arbitrary basis require solving a linear system to extract, its coordinates with respect to an orthonormal basis are handed to you directly: the $i$-th coordinate of $v$ is simply $\langle v, e_i\rangle$. This is why orthonormal bases are the preferred coordinate systems whenever an inner product is present — they convert the abstract inner product structure into concrete, computable formulas. The existence proof is not merely an abstract compactness or dimension-counting argument. It is completely constructive: start with any basis, and mechanically straighten it out, one vector at a time, using the Gram-Schmidt process. Each step subtracts off the components of a new vector that lie along previously-built orthonormal directions, leaving a vector orthogonal to all of them; then that leftover is rescaled to length one. Because the original list spanned $V$ and Gram-Schmidt preserves spans at each stage, the final orthonormal list still spans $V$, and having the right count of vectors, it is a basis.

Example 1: A genuinely skewed basis of $\mathbb{R}^3$. Let $V = \mathbb{R}^3$ with the standard inner product $\langle x,y\rangle = x_1y_1+x_2y_2+x_3y_3$. Consider the basis

(Check linear independence: the determinant of the matrix with these rows is $1(0\cdot1-1\cdot1) - 1(1\cdot1-1\cdot0)+0 = -1-1 = -2 \neq 0$, so they are independent, hence a basis of $\mathbb{R}^3$.)

Step 1. $\lVert v_1\rVert = \sqrt{1^2+1^2+0^2}=\sqrt2$, so

Step 2. Compute $\langle v_2,e_1\rangle = \frac{1}{\sqrt2}(1\cdot1+0\cdot1+1\cdot0) = \frac{1}{\sqrt2}$.

Then $\lVert f_2\rVert = \sqrt{\tfrac14+\tfrac14+1} = \sqrt{\tfrac32} = \frac{\sqrt6}{2}$, so

Step 3. Compute $\langle v_3,e_1\rangle = \frac{1}{\sqrt2}(0+1+0) = \frac{1}{\sqrt2}$, and $\langle v_3,e_2\rangle = \frac{1}{\sqrt6}(0-1+2) = \frac{1}{\sqrt6}$.

Compute componentwise:

Then $\lVert f_3\rVert = \sqrt{\tfrac49+\tfrac49+\tfrac49} = \sqrt{\tfrac{12}9} = \frac{2}{\sqrt3}$, so

Verification. Check $\langle e_1,e_3\rangle = \frac{1}{\sqrt6}(-1+1+0) = 0$, $\langle e_2,e_3\rangle = \frac{1}{\sqrt{18}}(-1-1+2)=0$, and each has norm $1$ (direct computation confirms this). So

is an orthonormal basis of $\mathbb{R}^3$, obtained from the skewed basis $v_1,v_2,v_3$, exactly as the theorem guarantees.

Example 2: Extending a single orthonormal vector in $\mathbb{C}^3$, and reading off coordinates. Let $V = \mathbb{C}^3$ with the standard Hermitian inner product $\langle x,y\rangle = x_1\overline{y_1}+x_2\overline{y_2}+x_3\overline{y_3}$, and suppose we are handed the single unit vector

Check $\lVert e_1\rVert^2 = \frac13(|1|^2+|i|^2+|1|^2) = \frac13(1+1+1)=1$. Good, it is a unit vector, so $\{e_1\}$ is an orthonormal list of length $1$ in the $3$-dimensional space $\mathbb{C}^3$. By the corollary, it extends to an orthonormal basis. First extend $e_1$ to any basis. Take $w_2 = (1,0,0)$, $w_3=(0,1,0)$; one checks $e_1, w_2, w_3$ is a basis (the matrix with these as rows is invertible: expanding along the third row, its determinant is $-1$ times the $2\times2$ minor from rows 1,2, columns 1,2, namely $\det\begin{pmatrix}1/\sqrt3 & i/\sqrt3\\1&0\end{pmatrix} = -i/\sqrt3 \ne 0$). Now Gram-Schmidt fixes $e_1$ (already unit) and processes $w_2$:

(Here the Hermitian inner product conjugates the second argument, so $\langle w_2,e_1\rangle = \sum_k (w_2)_k \overline{(e_1)_k}$.)

Then $\lVert f_2\rVert^2 = \tfrac49 + \tfrac19 + \tfrac19 = \tfrac69 = \tfrac23$, so $\lVert f_2\rVert = \sqrt{2/3}$, and

(Check: $\lVert e_2\rVert^2 = \frac16(4+1+1) = 1$. Good.) Next, $\langle w_3,e_1\rangle = 0 + 1\cdot\overline{i/\sqrt3} + 0 = -\frac{i}{\sqrt3}$, and

Compute the two correction terms: $\frac{i}{3}(1,i,1) = (\tfrac{i}3,-\tfrac13,\tfrac{i}3)$, and $\frac{i}{6}(2,-i,-1) = (\tfrac{i}3,\tfrac16,-\tfrac{i}6)$.

Then $\lVert f_3\rVert^2 = 0 + \tfrac14+\tfrac14 = \tfrac12$, $\lVert f_3\rVert = \tfrac{1}{\sqrt2}$, so

So

is an orthonormal basis of $\mathbb{C}^3$ extending the original vector. As a payoff, coordinates of any vector are now immediate: for instance the vector $u = (1,1,1)$ has

so the $e_1$-coefficient of $u$ in the expansion $u = \sum_k \langle u,e_k\rangle e_k$ is $\frac{2-i}{\sqrt3}$, obtained by a single inner product, with no linear system to solve.