Orthogonal and Orthonormal Matrices

April 13, 2026

Problem

Verify that Q = [[cos θ, −sin θ],[sin θ, cos θ]] is orthogonal by computing QᵀQ and QQᵀ. Show that Q preserves lengths and angles.

Explanation

Definition

A real square matrix QQ is orthogonal (the convention varies; some authors say "orthonormal") if its columns form an orthonormal set. Equivalently: QTQ=QQT=IQ^T Q = Q Q^T = I

which means Q1=QTQ^{-1} = Q^T — the inverse is just the transpose.

What the condition means, column by column

If QQ's columns are q1,q2,,qn\mathbf{q}_1, \mathbf{q}_2, \ldots, \mathbf{q}_n, then (QTQ)ij=qiqj(Q^T Q)_{ij} = \mathbf{q}_i \cdot \mathbf{q}_j. So QTQ=IQ^T Q = I is equivalent to

  • qiqj=0\mathbf{q}_i \cdot \mathbf{q}_j = 0 for iji \ne j (orthogonal columns), and
  • qiqi=1\mathbf{q}_i \cdot \mathbf{q}_i = 1 (unit length) — the columns are orthonormal.

Step-by-step — rotation matrix

Q=(cosθsinθsinθcosθ)Q = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

Step 1 — Compute QTQQ^T Q. QT=(cosθsinθsinθcosθ)Q^T = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}

QTQ=(cos2θ+sin2θcosθsinθ+sinθcosθsinθcosθ+cosθsinθsin2θ+cos2θ)Q^T Q = \begin{pmatrix} \cos^2\theta + \sin^2\theta & -\cos\theta \sin\theta + \sin\theta \cos\theta \\ -\sin\theta \cos\theta + \cos\theta \sin\theta & \sin^2\theta + \cos^2\theta \end{pmatrix}

=(1001)=I= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I ✓

Step 2 — Check QQTQ Q^T (should also equal II for square QQ).

By similar calculation, QQT=IQ Q^T = I ✓.

QQ is orthogonal.

Length preservation

For any x\mathbf{x}: Qx2=(Qx)T(Qx)=xTQTQx=xTx=x2\|Q \mathbf{x}\|^2 = (Q \mathbf{x})^T (Q \mathbf{x}) = \mathbf{x}^T Q^T Q \mathbf{x} = \mathbf{x}^T \mathbf{x} = \|\mathbf{x}\|^2

So Qx=x\|Q \mathbf{x}\| = \|\mathbf{x}\| — orthogonal matrices are isometries.

Angle preservation

(Qu)(Qv)=uTQTQv=uTv(Q \mathbf{u}) \cdot (Q \mathbf{v}) = \mathbf{u}^T Q^T Q \mathbf{v} = \mathbf{u}^T \mathbf{v}

The dot product is preserved — hence angles between vectors are preserved.

Together, length and angle preservation mean orthogonal matrices are rigid motions fixing the origin: rotations and reflections.

Orthogonal = rotation or reflection

A real orthogonal matrix always has detQ=±1\det Q = \pm 1:

  • detQ=+1\det Q = +1: QQ is a rotation (preserves orientation). These form SO(n)SO(n).
  • detQ=1\det Q = -1: QQ is a reflection (reverses orientation). Together with rotations they form O(n)O(n).

Key properties

  • Inverse is transpose: Q1=QTQ^{-1} = Q^T. Cheap to compute.
  • Product of orthogonal is orthogonal: if P,QP, Q are orthogonal, so is PQPQ.
  • Eigenvalues lie on the unit circle (modulus 1): λ=1|\lambda| = 1 for every eigenvalue.
  • Determinant: detQ=1|\det Q| = 1.

Why orthogonal matrices are amazing

  • Numerically stable. In floating-point, errors don't amplify under orthogonal multiplication.
  • Easy inverse. Transposing is O(n2)O(n^2) vs. general inversion at O(n3)O(n^3).
  • Building block of decompositions. QR, SVD, and spectral theorem all use orthogonal/unitary matrices.
  • Physics: represent symmetries (rotations of coordinate frames, spin, etc.).
  • Computer graphics: rotations preserve shape; reflections mirror it.

Orthogonal vs. orthonormal matrix?

Terminology varies. Strictly:

  • Orthonormal matrix: columns are orthonormal (orthogonal and unit length). Same as our definition above.
  • "Orthogonal matrix": most linear algebra texts use this for the orthonormal case.
  • Some applied contexts use "orthogonal" loosely to mean just orthogonal (not unit-length) columns, but that's non-standard.

The rule of thumb: "orthogonal matrix" ⟹ QTQ=IQ^T Q = I.

Common mistakes

  • Checking only orthogonality, not unit length. A matrix with perpendicular but non-unit columns is not orthogonal.
  • Forgetting detQ=±1\det Q = \pm 1. If the determinant is not one of these, the matrix is not orthogonal.
  • Assuming all symmetric matrices are orthogonal. They're different: symmetric means AT=AA^T = A; orthogonal means ATA=IA^T A = I.

Try it in the visualization

Rotate θ\theta through 00 to 2π2\pi. Columns of QQ traverse the unit circle; QTQQ^T Q stays as identity (all four entries display 1,0,0,11, 0, 0, 1). A unit square rotates without distortion — confirming length and angle preservation.

Interactive Visualization

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