M4-T2 — Orthogonal and Orthonormal Bases

Learning objectives

After this lesson, you should be able to:

• Define an orthonormal basis and compute coordinates of a vector in it with dot products.
• Derive the coordinate formula c_i = q_i · v and the length identity ‖v‖² = Σ c_i².
• Write the basis as the columns of a matrix Q and show QᵀQ = I.
• Explain why orthonormal bases are numerically ideal and where robots use them.

Introduction and motivation

A robot's forward, left and up axes are perpendicular unit vectors. The principal axes of a point cloud are perpendicular unit vectors. Every time you convert a measurement between such frames, you do it with dot products, not with a solver. This lesson explains why: an orthonormal basis is the one coordinate system in which the answer to “how much of v lies along this axis?” is a single multiplication and addition, and in which errors stay the size they started.

The intuition

A basis is orthogonal if its vectors are mutually perpendicular and orthonormal if, in addition, each has length 1. Like the ordinary x, y, z axes, but possibly rotated.

To find how far a vector reaches along a basis direction, shine a light perpendicular to that direction: the shadow length is the coordinate. With unit basis vectors the shadow is exactly the dot product, and because the axes are perpendicular, the coordinates do not interfere with each other.

Geometric and physical interpretation

Geometric view. An orthonormal basis is a rotated (possibly reflected) copy of the standard axes. Using it changes how you describe a vector but not its length or the angles between vectors: only the point of view has turned. Physical view. A robot frame with forward, left and up axes is an orthonormal basis. The components of a velocity along them are its forward speed, lateral slip and vertical rate; the squared components add up to the squared speed, so energy-like quantities are the same in either description.

The mathematics

q₁, …, q_n

an orthonormal basis: q_i · q_j = 0 for i ≠ j and q_i · q_i = 1.

Q

the matrix whose columns are q₁, …, q_n.

c_i

the coordinate of v along q_i.

Orthonormal means this

qi · qj = 0  (i ≠ j),     qi · qi = 1          ⇔          Qᵀ Q = I

What it means. Perpendicular, unit-length columns. Stacking the pairwise dot products into a matrix, QᵀQ has the dot product q_i·q_j in entry (i, j): that is the identity matrix exactly when the basis is orthonormal.

Coordinates are dot products

v = c1 q1 + ... + cn qn,     ci = qi · v          i.e.   c = Qᵀ v

What it means. No equation solving is needed: each coordinate is the shadow of v on one basis vector. In matrix form the coordinate vector is just Qᵀv, because Q⁻¹ = Qᵀ.

Lengths are preserved

‖v‖² = c1² + c2² + ... + cn²          (Parseval)

What it means. The squared length of v equals the sum of the squared coordinates in any orthonormal basis: changing from one orthonormal basis to another cannot change a length.

Orthonormal basis: coordinates are dot products, c = Qᵀv, the inverse is the transpose, and lengths are preserved.

Step-by-step derivation: coordinates as dot products, and Parseval

Let q₁, …, qn be an orthonormal basis and write v = c₁q₁ + … + cnqn (such coordinates exist because a basis spans).

1. Dot both sides with qj: qj·v = c₁(qj·q₁) + … + cn(qj·qn).
2. By orthonormality every dot product qj·qi is 0 for i ≠ j and 1 for i = j. All terms vanish except cj.
3. Therefore cj = qj·v. No system of equations was solved.

The matrix form. Collect the basis as the columns of Q. Then v = Q c. The entry (i, j) of QᵀQ is qi·qj, which is 1 on the diagonal and 0 elsewhere, so QᵀQ = I. Multiplying v = Qc by Qᵀ gives Qᵀv = QᵀQ c = c: the coordinates are c = Qᵀv, confirming Q⁻¹ = Qᵀ for a square Q.

Parseval. Compute the squared length using the expansion:

‖v‖² = v·v = (Σi ci qi)·(Σj cj qj) = Σi Σj ci cj (qi·qj) = Σi ci²

What it means. Only the terms with i = j survive, each contributing c_i² times 1. The cross terms vanish because the basis vectors are orthogonal.

Equivalently, for a square Q: ‖c‖² = cᵀc = (Qᵀv)ᵀ(Qᵀv) = vᵀQQᵀv = vᵀv = ‖v‖², because QQᵀ = I for a square orthonormal Q. The coordinate vector has the same length as v.

A worked example: a rotated orthonormal basis

Let q₁ = (0.6, 0.8) and q₂ = (−0.8, 0.6), and v = (2, 1).

1. Orthonormal? q₁·q₁ = 0.36 + 0.64 = 1, q₂·q₂ = 0.64 + 0.36 = 1, q₁·q₂ = −0.48 + 0.48 = 0. ✓
2. Coordinates. c₁ = q₁·v = 1.2 + 0.8 = 2 and c₂ = q₂·v = −1.6 + 0.6 = −1.
3. Rebuild. 2 q₁ − q₂ = (1.2 + 0.8, 1.6 − 0.6) = (2, 1) = v. ✓
4. Parseval. ‖v‖² = 4 + 1 = 5 and c₁² + c₂² = 4 + 1 = 5. ✓ The lengths agree.
5. Matrix form. Q = [[0.6, −0.8], [0.8, 0.6]] and QᵀQ = I; Qᵀv = (0.6·2 + 0.8·1, −0.8·2 + 0.6·1) = (2, −1).

Contrast with a skewed basis. For b₁ = (1, 1), b₂ = (−1, 2) (the earlier lesson) the coordinates of (3, 0) were (2, −1) but required solving a system, and the dot products b₁·v = 3 and b₂·v = −3 are not the coordinates.

Robotics application

Robot frames. A frame's three axes form an orthonormal basis, so converting a velocity or point into the frame is multiplication by Qᵀ (Module 7: a rotation matrix). That is why the inverse of a rotation is cheap: it is the transpose.

Principal axes. The directions of greatest and least spread of a point cloud, found with PCA (Module 6), are orthonormal. Expressing the cloud in them removes correlations and makes a flat object (a wall, the ground) one coordinate thin.

Numerical stability. Computations in an orthonormal basis do not magnify errors: ‖Qx‖ = ‖x‖, so any error in x reaches the output at the same size. Algorithms built on orthonormal factors (QR, SVD) are the most reliable in numerical linear algebra, an advantage Module 5 makes concrete.

Sensor fusion. Decomposing an acceleration into components along known orthogonal axes (gravity direction and the horizontal plane) is how attitude is estimated from an accelerometer; see attitude representation and rotation matrices.

Engineering implications and limitations

Practical rules

• Use Qᵀ, not Q⁻¹. For an orthonormal basis the transpose is the exact inverse. Calling a general inverse is slower and less accurate.
• Check orthonormality numerically. After constructing a basis, verify max|QᵀQ − I| is at the level of machine precision (about 10⁻¹⁵). A value of 10⁻³ means your frame has drifted or been mis-specified.
• Normalise first. The dot-product formula assumes unit-length basis vectors. For an orthogonal but unnormalised basis use ci = (v·bi)/(bi·bi).
• Re-orthonormalise after updates. Accumulated rounding degrades orthonormality (see Gram-Schmidt and QR next); do it periodically for integrated rotations.
• Mind handedness. An orthonormal basis can be right- or left-handed; the determinant of Q is +1 or −1. Frames used for rotations must be right-handed (det +1).

Limits

Orthonormal bases are not always available in the coordinates you were given: a camera pixel grid is not square in metres, and a set of joint angles is not an orthonormal basis of hand motion. You then use a general basis and a solve, or convert to an orthonormal one first.

Common misconceptions

Engineering takeaways

1. An orthonormal basis has mutually perpendicular unit vectors: QᵀQ = I.
2. Coordinates are dot products, c_i = q_i·v, or c = Qᵀv in matrix form.
3. Parseval: ‖v‖² = Σ c_i², so changing orthonormal basis never changes length.
4. For a square Q, the inverse is the transpose, the cheapest and most accurate inverse there is.
5. Robot frames and principal axes are orthonormal bases, which is why they are used so widely.

References and further reading

Primary papers and authoritative documentation.

• MIT 18.06SC Linear Algebra (OCW): lectures on orthogonal bases and Gram-Schmidt
• Mathematics for Machine Learning: the chapter on analytic geometry: orthonormal bases
• Modern Robotics (Lynch and Park), free preprint: chapter 3: rotation matrices as orthonormal frames
• Immersive Linear Algebra: chapters on the dot product and linear mappings

Knowledge check

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What’s next

Orthonormal bases make decomposition easy. The next lesson uses this machinery to answer a different question: what is the closest point to a given vector inside a line or plane? That is projection.

Related content

Nothing below is required to finish this lesson.

• Before
  M4-T1: Orthogonal Vectors

  Test orthogonality with the dot product, split a vector into orthogonal parts, and use right angles in robot geometry.