datarekha

Orthogonality & Orthogonal Matrices

Orthogonal matrices have orthonormal columns, so Q transpose times Q equals the identity. They preserve length and angle, have determinant plus or minus one, and eigenvalues on the unit circle.

8 min read Advanced GATE DA Lesson 27 of 122

What you'll learn

  • Two vectors are orthogonal when their dot product is zero
  • An orthogonal matrix Q has orthonormal columns, so Q transpose times Q = I and the inverse equals the transpose
  • Orthogonal matrices preserve length and angle: the norm of Qx equals the norm of x
  • det(Q) = plus or minus 1, and every eigenvalue lies on the unit circle (so it may be complex, not just plus or minus 1)

Before you start

Picture turning a piece of paper on the desk. The shapes drawn on it shift, but nothing stretches and no right angle bends. That’s an orthogonal matrix doing its job — it moves vectors around without ever changing a length or an angle. The formal name for that family is “rigid motions”: rotations and reflections, nothing fancier. GATE loves the topic because preserving every distance forces a whole cascade of clean properties.

Start with the word in isolation: two vectors are orthogonal when their dot product is zero, u · v = 0 — they meet at a right angle. An orthogonal matrix is the natural upgrade: a whole basis of directions that are mutually perpendicular and unit-length.

Orthonormal columns and Q transpose Q = I

A square matrix Q is orthogonal when its columns are orthonormal — each column is a unit vector, and any two different columns are orthogonal. Stack that condition into matrix form and it collapses to one clean equation:

Qᵀ Q = IsoQ⁻¹ = Qᵀcolumn i dotted with column j = 1 if i = j, else 0
Entry (i, j) of Q transpose times Q is column i dotted with column j — which is exactly the identity.

Because QᵀQ = I, the inverse is simply the transpose: Q⁻¹ = Qᵀ. Inverting an orthogonal matrix costs nothing — you just flip it. (The same holds row-wise: the rows are orthonormal too, so QQᵀ = I as well.)

They preserve length and angle

The defining geometric fact: an orthogonal matrix never changes the length of a vector. For every x,

‖Qx‖² = (Qx)ᵀ(Qx) = xᵀ QᵀQ x = xᵀ I x = xᵀx = ‖x‖²

so ‖Qx‖ = ‖x‖. The same QᵀQ = I cancellation shows dot products survive too ((Qx)·(Qy) = x·y), so angles are preserved. A rotation is the clearest picture — the vector spins, its length is untouched:

xQxθboth vectors touch the same dashed circle: equal length
A rotation moves x to Qx along a circle of fixed radius — the norm is preserved.

Try the slider feel for yourself. In the playground below, set the matrix entries to cosθ, −sinθ, sinθ, cosθ (an honest rotation) and the unit shape spins without distorting. Drag the entries off those rotation values and lengths and angles change immediately — that’s the visual signature of not being orthogonal.

Determinant and eigenvalues

Take determinants of QᵀQ = I: since det(Qᵀ) = det(Q), we get det(Q)² = 1, so det(Q) = +1 or −1. A +1 is a pure rotation; a −1 includes a reflection.

The eigenvalues are subtler. Because lengths are preserved, if Qx = λx then ‖x‖ = ‖Qx‖ = |λ|·‖x‖, forcing |λ| = 1. Every eigenvalue sits on the unit circle in the complex plane. That is not the same as saying every eigenvalue is ±1 — a genuine rotation has no real eigenvectors at all, and its eigenvalues are a complex conjugate pair cosθ ± i·sinθ.

How GATE asks this

This is an MSQ favourite: you are told a real matrix A satisfies ‖Ax‖ = ‖x‖ for all x, and asked which conclusions follow. The trustworthy ones — A is orthogonal, A is full rank (invertible), det(A) = ±1 — are mixed with the trap option “the eigenvalues of A are ±1,” which is false. GATE DA 2025 ran exactly this, baiting candidates who memorised “orthogonal” without remembering that rotations have complex eigenvalues.

Worked example — the 2025 norm-preserving question

Let A be a real n × n matrix with ‖Ax‖ = ‖x‖ for every x in Rⁿ. Which of the following must be true?

Step 1 — orthogonal. Preserving every length preserves every dot product, hence AᵀA = I. So A is orthogonal. TRUE.

Step 2 — full rank. AᵀA = I means A has an inverse (namely Aᵀ), so A is invertible and full rank. TRUE.

Step 3 — determinant. det(A)² = det(AᵀA) = det(I) = 1, so det(A) = ±1. TRUE.

Step 4 — the trap. “All eigenvalues are ±1.” FALSE. Take the 2-D rotation by θ = 90°,

A = [ 0  −1 ]      AᵀA = [ 1  0 ] = I,   det A = 1,
    [ 1   0 ]            [ 0  1 ]

eigenvalues of A:  λ = ± i   (|λ| = 1, but complex, not ±1)

A is orthogonal and full rank, yet its eigenvalues are +i and −i — on the unit circle, but neither +1 nor −1. The correct selections are orthogonal, full rank, and det = ±1; the eigenvalue option is the distractor.

For the second part — verifying a general rotation — expand QᵀQ for Q = [[cosθ, −sinθ], [sinθ, cosθ]]:

QᵀQ = [ cos²θ + sin²θ        −cosθ·sinθ + sinθ·cosθ ]   = [ 1  0 ]
      [ −sinθ·cosθ + cosθ·sinθ   sin²θ + cos²θ      ]     [ 0  1 ]

det Q = cosθ·cosθ − (−sinθ)·sinθ = cos²θ + sin²θ = 1

so every rotation matrix is orthogonal with determinant +1.

Quick check

Quick check

0/6
Q1Q is a 3 by 3 orthogonal matrix. What are the only possible values of det(Q)? (select all that apply)select all that apply
Q2Q is orthogonal and x is a vector with ‖x‖ = 5. What is ‖Qx‖?numerical answer — type a number
Q3A real matrix A satisfies ‖Ax‖ = ‖x‖ for all x. Which statements MUST be true? (select all that apply)select all that apply
Q4Which 2 by 2 matrix is orthogonal? (select all that apply)select all that apply
Q5The eigenvalues of a 2-D rotation matrix by angle θ = 60° are cos60° ± i·sin60°. What is the modulus |λ| of each eigenvalue?numerical answer — type a number
Q6If Q is orthogonal, which equalities hold? (select all that apply)select all that apply

Sign in to track your progress

Completed lessons, your XP, level, and streak save to your account — it's free and takes a few seconds.

Related lessons

Skip to content