Questions: Orthonormal Bases

Question 1

A square matrix Q has columns that are pairwise orthogonal (each pair is perpendicular). A student concludes that Q is an orthogonal matrix, so Qᵀ = Q⁻¹. Is the student correct?

• A. Yes — pairwise orthogonality of columns is sufficient for Qᵀ = Q⁻¹
• B. No — the columns must also have unit length; without normalization, QᵀQ ≠ I
• C. No — Q must also be symmetric for the property to hold
• D. Yes — for square matrices, orthogonal columns always satisfy QᵀQ = I

Answer: (B)

The term 'orthogonal matrix' is confusingly named: it requires orthonormal columns, not merely orthogonal ones. QᵀQ = I requires that columns have unit length AND are pairwise perpendicular. If columns are perpendicular but not unit length, QᵀQ is a diagonal matrix with the squared norms on the diagonal — not the identity. The student identified one of the two required conditions and missed the other.

Question 2

You are working in ℝⁿ with an orthonormal basis {u₁, u₂, ..., uₙ}. How do you find the coordinate of a vector v with respect to u₃?

• A. Solve the linear system Uc = v for c, then take the third component
• B. Compute ⟨v, u₃⟩ — the inner product of v with u₃
• C. Project v onto the span of {u₁, u₂} and subtract from v, then normalize
• D. Compute ‖v‖ / ‖u₃‖

Answer: (B)

This is the defining computational advantage of an orthonormal basis. For a general basis, finding coordinates requires solving a linear system. For an ONB, the coordinate with respect to uᵢ is simply ⟨v, uᵢ⟩. This works because when you expand v = c₁u₁ + ... + cₙuₙ and take the inner product with u₃, all terms vanish by orthogonality except ⟨c₃u₃, u₃⟩ = c₃·1 = c₃. The unit-length condition is what eliminates the denominator.

Question 3

If Q is an orthogonal matrix, then the transformation v ↦ Qv preserves both lengths and angles.

• T. True
• F. False

Answer: True

Orthogonal matrices preserve the inner product: ⟨Qu, Qv⟩ = ⟨u, v⟩. Since lengths are determined by ‖v‖ = √⟨v,v⟩ and angles by cos θ = ⟨u,v⟩/(‖u‖‖v‖), preserving inner products implies preserving both. This is why rotations and reflections are orthogonal matrices — they are precisely the rigid motions of space.

Question 4

Any set of nonzero pairwise-orthogonal vectors forms an orthonormal basis for its span.

• T. True
• F. False

Answer: False

An orthonormal basis requires three properties: (1) pairwise orthogonality, (2) unit length for every vector, and (3) spanning. A set of pairwise-orthogonal nonzero vectors satisfies (1) and (3) but not necessarily (2). For example, {(2,0), (0,3)} is pairwise orthogonal but neither vector has unit length, so it is not orthonormal. Dividing each vector by its norm gives the orthonormal version {(1,0), (0,1)}.

Question 5

Why does computing coordinates in an orthonormal basis reduce to taking dot products, whereas in a general basis it requires solving a linear system?

Model answer: In a general basis, the basis vectors are not perpendicular, so when you expand v = c₁b₁ + ... + cₙbₙ and take the inner product with one basis vector bᵢ, the cross-terms ⟨cⱼbⱼ, bᵢ⟩ for j ≠ i don't vanish — they produce a system of equations coupling all coordinates simultaneously. In an ONB, ⟨uᵢ, uⱼ⟩ = 0 for i ≠ j, so every cross-term drops out, leaving ⟨v, uᵢ⟩ = cᵢ directly. Orthogonality decouples the coordinates; normalization removes the denominators.

The key structure is the Gram matrix: for a general basis, coordinates require inverting [⟨bᵢ,bⱼ⟩]ᵢⱼ. For an ONB, the Gram matrix is the identity, so the coordinates are just dot products. This decoupling is the computational miracle that makes orthonormal bases the preferred choice whenever flexibility of basis choice exists.