Orthogonal sets and orthonormal basis: Question bank

True or false — justify

Every claim is a statement you must label true or false — but the score is in the justification.

An orthogonal set is always linearly independent.

False — only nonzero orthogonal sets are. The zero vector satisfies

0 \cdot u_{j} = 0

vacuously, so

{0, u}

is "orthogonal" yet dependent.

u \cdot v = 0

then

{u, v}

is orthonormal.

False — orthogonal only tells you they are perpendicular; orthonormal additionally needs

∥ u ∥ = ∥ v ∥ = 1

, which the dot product being zero says nothing about.

Every orthogonal set of

n

nonzero vectors in

R^{n}

is a basis of

R^{n}

True — they are automatically independent, and

n

independent vectors in

R^{n}

always span it, hence form a basis.

A basis of

R^{3}

can contain 4 mutually orthogonal nonzero vectors.

False — 4 nonzero orthogonal vectors are independent, but

R^{3}

holds at most 3 independent vectors, so no such set exists at all.

U

has orthonormal columns then

U

is invertible.

False in general —

U^{T} U = I

holds, but if

U

m \times p

with

m > p

it is not square and has no inverse; only square orthonormal-column matrices are invertible.

For any matrix

U

with orthonormal columns,

U U^{T} = I

False unless

U

is square.

U U^{T}

is the projection onto the column space of

U

; it equals

I

only when those columns span all of

R^{m}

Multiplying a vector by an orthogonal matrix can change its length.

False —

∥ U x ∥ = ∥ x ∥

for orthonormal columns, since

∥ U x ∥^{2} = x^{T} U^{T} U x = x^{T} x

An orthogonal matrix has determinant

+ 1

False — orthogonal matrices satisfy

det (U) = \pm 1

;

- 1

corresponds to reflections (an odd number of them),

+ 1

to pure rotations.

If a set is orthonormal, dividing each vector by its length leaves it unchanged.

True — orthonormal vectors already have length 1, so normalizing does nothing; this is why orthonormal is the "already-finished" state.

Coordinates in an orthonormal basis require solving a linear system.

False — that is the whole payoff:

c_{j} = y \cdot u_{j}

is a single dot product per coordinate, no elimination needed.

Spot the error

Each line contains a flawed statement; the reveal names the mistake and repairs it.

{u_{1}, u_{2}}

is orthogonal, so its coordinate formula is

c_{j} = y \cdot u_{j}

The denominator was dropped — for a merely orthogonal (not normalized) basis the formula is

c_{j} = \frac{y \cdot u _{j}}{u _{j} \cdot u _{j}}

; the clean version needs unit vectors.

"The columns of

U

are orthogonal, so

U^{T} U = I

Only orthogonal, not orthonormal — then

U^{T} U

is a diagonal matrix with entries

∥ u_{j} ∥^{2}

, equal to

I

only after normalizing.

U

is orthogonal, so its rows might not be orthonormal."

For a square orthogonal matrix

U^{T} U = U U^{T} = I

, so the rows are orthonormal too; row- and column-orthonormality are equivalent for square

U

"This orthogonal set has 5 vectors including

0

, and it's independent because they're all perpendicular."

The zero vector is perpendicular to everything but contributes a nonzero coefficient in

0 \cdot u + 0 = 0

, breaking independence — the "nonzero" hypothesis is essential.

"To project

y

onto

W

I need

W

's basis to be orthonormal."

You only need it orthogonal; the Orthogonal projection formula

\sum \frac{y \cdot u _{j}}{u _{j} \cdot u _{j}} u_{j}

works for any orthogonal basis, with the denominator handling non-unit lengths.

"An orthonormal set of 2 vectors in

R^{3}

is an orthonormal basis."

It's an orthonormal set, but a basis of

R^{3}

needs 3 vectors; it is only a basis of the 2-dimensional subspace it spans.

"Since

U^{T} U = I

, we have

U^{- 1} = U^{T}

This inference is valid only when

U

is square; for tall

U

U^{T}

is a left inverse, not a genuine inverse.

Why questions

These probe the reason behind each fact, not the fact itself.

Why does dotting the dependence relation

\sum c_{i} u_{i} = 0

with

u_{j}

prove independence?

Orthogonality kills every cross term

u_{i} \cdot u_{j}

(

i \neq = j

), leaving only

c_{j} ∥ u_{j} ∥^{2} = 0

; since

∥ u_{j} ∥^{2} > 0

for nonzero vectors,

c_{j} = 0

for every

j

Why is the term "orthogonal matrix" a historical trap?

It requires orthonormal columns and squareness, not merely orthogonal ones — the name kept "orthogonal" even though unit length and squareness are baked in.

Why do orthonormal-column maps preserve dot products, and why does that matter numerically?

(U x) \cdot (U y) = x^{T} U^{T} U y = x \cdot y

, so angles and lengths are untouched — such maps never amplify errors, which is why QR decomposition is numerically stable.

Why is

u_{j} \cdot u_{j}

in the coordinate denominator, not

∥ u_{j} ∥

The isolated term after dotting is

c_{j} (u_{j} \cdot u_{j}) = c_{j} ∥ u_{j} ∥^{2}

, so dividing by

u_{j} \cdot u_{j} = ∥ u_{j} ∥^{2}

(the squared length) is exactly what cancels it.

Why does an orthonormal basis make Least squares easy?

The normal-equation matrix

U^{T} U

becomes

I

, so the least-squares solution collapses to

\hat{x} = U^{T} b

— dot products instead of solving a system.

Why does Gram-Schmidt process exist if orthonormal bases are so convenient?

Because most given bases are not orthogonal; Gram-Schmidt manufactures an orthonormal basis from an arbitrary one so you can then enjoy the dot-product shortcuts.

Why can two eigenvectors of a symmetric matrix be chosen orthonormal?

Eigenvalues and eigenvectors of a symmetric matrix from distinct eigenvalues are automatically orthogonal, and normalizing them yields an orthonormal eigenbasis — the spectral theorem's payoff.

Edge cases

Degenerate, zero, and limiting inputs the definitions quietly cover.

Is the empty set orthogonal?

Vacuously yes — there are no distinct pairs to violate

u_{i} \cdot u_{j} = 0

, so it satisfies the definition trivially (and is independent).

Is a single nonzero vector an orthogonal set?

Yes — there is no distinct pair to check, so any lone nonzero vector counts as orthogonal, and it becomes orthonormal after dividing by its length.

Can a unit vector be part of an orthogonal-but-not-orthonormal set?

Yes — orthonormality demands all vectors have length 1; a set with one unit vector and one length-3 vector is orthogonal but not orthonormal.

What is the coordinate of

y = 0

in any orthogonal basis?

Every

c_{j} = \frac{0 \cdot u _{j}}{u _{j} \cdot u _{j}} = 0

, so the zero vector has all-zero coordinates, as expected.

y

lies exactly along

u_{2}

, what do its coordinates look like?

Only

c_{2}

is nonzero and all other

c_{j} = 0

, because

y \cdot u_{j} = 0

for the perpendicular directions — a clean illustration of decoupling.

What happens to the coordinate formula if one basis vector is

0

It breaks —

u_{j} \cdot u_{j} = 0

makes the denominator zero, which is exactly why the zero vector is forbidden from a basis.

Is the

1 \times 1

matrix

[- 1]

an orthogonal matrix?

Yes — its single column has length 1 and

(- 1) (- 1) = 1 = I_{1}

; it represents the reflection

x \mapsto - x

and has determinant

- 1

R^{n}

, what is the largest possible orthonormal set?

Exactly

n

vectors — any more would be independent beyond the dimension, which is impossible, so a maximal orthonormal set is always an orthonormal basis.

Recall One-line summary of the traps

Orthogonal ≠ orthonormal ≠ orthogonal-matrix; the zero vector wrecks independence; tall orthonormal-column matrices give $U^{T} U = I$ but not $U U^{T} = I$ ; and never drop the $u_{j} \cdot u_{j}$ denominator unless lengths are truly 1.