Every Finite-Dimensional Inner Product Space Has an Orthonormal Basis

From an Arbitrary Basis to an Orthonormal One

An orthonormal basis combines two useful features: its vectors describe every vector in the space, and distinct basis vectors are mutually orthogonal and have length one. In a finite-dimensional inner product space, such a basis is not merely a convenient possibility. Starting with any basis, the Gram-Schmidt process constructs one. The key is that each new input vector has its earlier components removed; independence ensures that something nonzero remains to normalize.

Work over \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). An inner product on a vector space \(V\) is a positive-definite inner product, with the convention that it is linear in its first slot and conjugate-linear in its second slot when \(\mathbb{F}=\mathbb{C}\). Write \(\|\mathbf{v}\|=\sqrt{\langle\mathbf{v},\mathbf{v}\rangle}\). A list is orthonormal if each vector has norm one and distinct vectors have inner product zero. A basis is an ordered, linearly independent list that spans \(V\).

The coefficient in the recursion is chosen so that the corresponding component of \(\mathbf{v}_k\) is removed. The denominator is \(\langle\mathbf{u}_j,\mathbf{u}_j\rangle=\|\mathbf{u}_j\|^2\), which is positive for a nonzero residual. Thus the construction is defined as long as its residuals do not vanish. The next result establishes both that they do not vanish for independent inputs and that the construction preserves the span.

Why the Construction Works

Proof. We proceed in order through the list. For \(k=1\), independence gives \(\mathbf{v}_1\ne\mathbf{0}\), so \(\mathbf{u}_1=\mathbf{v}_1\ne\mathbf{0}\). Normalizing it gives \(\|\mathbf{q}_1\|=1\), and the first spans agree.

Suppose the construction has reached step \(k\), with \(\mathbf{q}_1,\dots,\mathbf{q}_{k-1}\) orthonormal and with the first \(k-1\) spans equal. The recursion gives \(\mathbf{u}_k\). For each \(i<k\), using linearity in the first slot and orthonormality,

Here \(\mathbf{u}_j=\|\mathbf{u}_j\|\mathbf{q}_j\), so the \(j=i\) term in the sum equals \(\langle\mathbf{v}_k,\mathbf{q}_i\rangle\), and every other term is zero. Hence \(\mathbf{u}_k\) is orthogonal to each earlier output.

If \(\mathbf{u}_k=\mathbf{0}\), the recursion would express \(\mathbf{v}_k\) as a linear combination of \(\mathbf{u}_1,\dots,\mathbf{u}_{k-1}\), and therefore of \(\mathbf{q}_1,\dots,\mathbf{q}_{k-1}\). By the span equality at step \(k-1\), it would then be a linear combination of \(\mathbf{v}_1,\dots,\mathbf{v}_{k-1}\), contradicting independence of the input list. Thus \(\mathbf{u}_k\ne\mathbf{0}\), so \(\mathbf{q}_k\) is defined and has norm one. It is orthogonal to every earlier \(\mathbf{q}_i\).

Finally, the recursion expresses \(\mathbf{u}_k\) as \(\mathbf{v}_k\) minus a combination of earlier vectors. Thus \(\mathbf{u}_k\), and hence \(\mathbf{q}_k\), belongs to the span of \(\mathbf{v}_1,\dots,\mathbf{v}_k\). Conversely, rearranging the recursion expresses \(\mathbf{v}_k\) as \(\mathbf{u}_k\) plus a combination of earlier residuals. Since each earlier residual is a nonzero scalar multiple of its normalized output, the two spans through step \(k\) are equal. This completes the induction. \(\square\)

Proof. If \(V\) has dimension zero, its empty basis is orthonormal: there are no vectors whose norms or pairwise inner products need to be checked. Otherwise, since \(V\) is finite-dimensional, it has a basis \((\mathbf{v}_1,\dots,\mathbf{v}_n)\) for some positive integer \(n\). A basis is linearly independent, so the proposition applies. It produces an orthonormal list with the same span as the basis. That span is \(V\), so the output list is an orthonormal basis of \(V\). \(\square\)

The theorem asserts existence, not that a particular orthonormal basis is unique. Different starting bases, or a different ordering of the same basis, can produce different outputs. What is guaranteed is that the process yields \(n\) orthonormal vectors spanning the same \(n\)-dimensional space.

Worked Constructions

A Common Mistake: The Projection Denominator

In the \(\mathbb{R}^2\) example, a tempting but incorrect coefficient for removing the component along \(\mathbf{u}_1=(1,1)^{\top}\) is \(\langle\mathbf{v}_2,\mathbf{u}_1\rangle/\|\mathbf{u}_1\|\). That would give \(1/\sqrt{2}\), and the supposed residual would be \((1-1/\sqrt{2},-1/\sqrt{2})^{\top}\). Its dot product with \(\mathbf{u}_1\) is \(1-\sqrt{2}\ne0\), so the component was not removed.

The correct coefficient is \(\langle\mathbf{v}_2,\mathbf{u}_1\rangle/\langle\mathbf{u}_1,\mathbf{u}_1\rangle=1/2\), not a division by the norm. It gives residual \((1/2,-1/2)^{\top}\), whose dot product with \(\mathbf{u}_1\) is \(1/2-1/2=0\). Keep the squared norm in the projection formula; normalization by the norm happens afterward.

What the Theorem Provides

For any basis of a finite-dimensional inner product space, Gram-Schmidt preserves the span of each initial segment while replacing the list by an orthonormal one. Independence is essential: if an input is already in the span of the earlier inputs, its residual is zero and cannot be normalized. For a basis, that case cannot occur. The resulting orthonormal basis supports simpler coordinate calculations because inner products with its vectors directly identify coefficients: if \(\mathbf{v}=\sum_{j=1}^n c_j\mathbf{q}_j\), then \(\langle\mathbf{v},\mathbf{q}_i\rangle=c_i\). This follows by taking the inner product of the sum with \(\mathbf{q}_i\) and using orthonormality.