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.
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:
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:
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
Abe a realn × nmatrix with‖Ax‖ = ‖x‖for everyxinRⁿ. 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.