QR Decomposition: Methods & Applications

What QR Decomposition Is

An

m \times n

matrix

A

with

m \geq n

and linearly independent columns factors as

QR Decomposition

A = QR

Qr decomposition →

where

Q

is

m \times n

with orthonormal columns and

R

is

n \times n

upper triangular with positive diagonal entries.

COROLLARY

The columns of

Q

form an orthonormal basis for the column space of

A

.

The matrix

R

stores the coefficients: each column of

A

is a linear combination of the columns of

Q

with weights given by the corresponding column of

R

. The upper triangular structure of

R

reflects the sequential nature of the orthogonalization — each column depends only on the columns that came before it.

A = QR, complete

The finished factorization: Q holds orthonormal columns spanning the same space as the columns of A, and R is upper triangular, recording how much of each orthonormal direction every original column uses. Because each column of A only involves the directions found up to that point, R has zeros below its diagonal. Build it column by column on the QR decomposition visualizer.

Q carries the geometry and R carries the bookkeeping; together they rebuild A exactly.

QR Notation

Notation

QR Notation

The same two letters appear on this page with three different jobs, and the shapes behind them are never written down. Five marks decide which reading applies.

The contract

A = QR

itself — what each factor carries and why

R

comes out upper triangular — is set out at the Gram-Schmidt process. The reserved letter

Q

and the equation

Q^{T}Q = I

belong to orthogonal sets, the normal equations and the hat on

\hat{\mathbf{x}}

to least squares.

A = Q_1R_1 \qquad \text{versus} \qquad A = QR

A equals Q one R one — the thin factorisation — against A equals Q R, the full one

The letters match and the matrices do not. Thin:

Q_1

is

m \times n

with orthonormal columns and

R_1

is

n \times n

. Full:

Q

is square

m \times m

and

R

is

m \times n

, carrying

m - n

rows of zeros underneath. The line

A = QR

records neither shape, so which version is meant has to come from the surrounding sentence.

CasesThe subscript

1

marks the thin factors only when both versions are in play; on their own the thin factors are usually written

Q

and

R

as well. Gram-Schmidt produces the thin version, and the full one is reached by appending

m - n

columns that span the orthogonal complement of the column space.

Also writtenReduced and economy are the same thing as thin. Numerical libraries select between them with an argument rather than a symbol, which is why code and page can disagree about what

Q

names.

Do not confuseAssuming

Q

is square. That assumption is harmless for the full version and is exactly where the identity in the next entry stops holding.

Q^{T}Q = I_n, \qquad QQ^{T} \neq I_m

Q transpose Q is the n by n identity; Q Q transpose is not the identity

For a thin

Q

the order of the product decides whether the statement is true. Orthonormal columns give

Q^{T}Q = I_n

directly. Reversing the factors gives

QQ^{T}

, the projection onto the column space of

A

— equal to the identity only when

Q

is square and there is nothing left over to project away.

CasesIn the full factorisation

Q

is square and both products give

I_m

, which is the case set out at orthogonal sets. This page needs the other one, and that is why

QQ^{T}

appears here as a projection matrix rather than as a way of writing

I

.

Also writtenThe thin case is sometimes flagged in words as orthonormal columns rather than by the equation, precisely to avoid claiming more than holds.

Do not confuseCalling a thin

Q

an orthogonal matrix. That name requires a square matrix, and a thin

Q

has no inverse at all. The practical version of the error is cancelling

QQ^{T}

inside a longer expression as though it were

I

.

R_{jj} = \|\mathbf{u}_j\| > 0

the j-th diagonal entry of R is the norm of the j-th orthogonal vector, and it is positive

Positivity buys uniqueness and is a convention, not something the factorisation forces. Multiplying a column of

Q

by

-1

and the matching row of

R

by

-1

leaves the product untouched, so without the sign rule an

n

-column matrix has

2^n

different QR factorisations. The line

A = QR

records none of that.

CasesGram-Schmidt gets the positive diagonal for free, since each

R_{jj}

is a norm. Householder reflections do not, and numerical libraries generally do not enforce the convention — two correct computed factorisations of one matrix can differ by a column of sign flips with nothing wrong in either.

Also writtenStated as a condition on

R

rather than built into the notation: with the diagonal of

R

kept positive, appended to the factorisation whenever uniqueness is being claimed.

Do not confuseReading the QR factorisation as though the article were earned. It is earned only once the sign convention is stated, in the same way that the diagonal of ones earns uniqueness at LU.

A_k = Q_kR_k, \qquad A_{k+1} = R_kQ_k

factor A sub k into Q sub k times R sub k, then multiply those same factors back in the opposite order

The letters

Q

and

R

name a factorisation everywhere else on this page and an iteration here. The whole method sits in the reversal: the second line multiplies the factors just produced in swapped order, which amounts to

A_{k+1} = Q_k^{T}A_kQ_k

— a similarity transformation, so the eigenvalues survive while the sub-diagonal shrinks.

CasesThe subscript

k

counts iteration steps. It is the only subscript on the page that indexes time rather than a column, a row or an entry, and

Q_k

is a different matrix at every step.

Also writtenThe shifted form factors

A_k - \sigma_kI

and adds

\sigma_kI

back afterwards, keeping the same two-line shape with a scalar moved in and out.

Do not confuseReading

R_kQ_k

as a second factorisation, or expecting the reversal to undo the first line. Swapping the order changes the matrix, and repeating the swap is what drives

A_k

toward upper triangular form with the eigenvalues on its diagonal.

H = I - \frac{2\mathbf{v}\mathbf{v}^{T}}{\mathbf{v}^{T}\mathbf{v}}

H equals the identity minus two times v v transpose, over v transpose v

One vector appears four times in two products that are not the same operation.

\mathbf{v}\mathbf{v}^{T}

is an

m \times m

matrix of rank one;

\mathbf{v}^{T}\mathbf{v}

is a single number, the squared length. Only the position of the transpose separates them, and it determines the shape of the result.

CasesThe numerator is a matrix and the denominator a scalar, so the fraction is ordinary scaling of a matrix and the subtraction from

I

is well formed. Writing

\mathbf{v}

as a unit vector makes the denominator

1

and shortens the formula to

I - 2\mathbf{v}\mathbf{v}^{T}

, the form most texts quote — the same reflection either way.

Also writtenThe rank-one term is sometimes written

\mathbf{v} \otimes \mathbf{v}

to name the outer product explicitly instead of leaving it to the transpose position.

Do not confuseReading the two products as interchangeable. Swapping them makes the subtraction ill-formed in one direction and the division ill-formed in the other, which is a useful check when the formula is being recalled rather than copied.

All linear algebra symbols →↑ Matrix decompositions

QR via Gram-Schmidt

Applying the Gram-Schmidt process to the columns

\mathbf{a}_1, \dots, \mathbf{a}_n

of

A

produces orthonormal vectors

\mathbf{q}_1, \dots, \mathbf{q}_n

. These become the columns of

Q

.

The entries of

R

are the dot products computed during Gram-Schmidt:

QR Gram-Schmidt R Entries

R_{ij} = \mathbf{q}_i \cdot \mathbf{a}_j \;\; (i \leq j), \qquad R_{ij} = 0 \;\; (i > j), \qquad R_{jj} = \|\mathbf{u}_j\|

Qr gram schmidt r entries →

Each column of

A

decomposes as

QR via Gram-Schmidt

\mathbf{a}_j = R_{1j}\mathbf{q}_1 + R_{2j}\mathbf{q}_2 + \cdots + R_{jj}\mathbf{q}_j

The diagonal entry

R_{jj} = \|\mathbf{u}_j\|

— the norm of the

j

-th orthogonal vector before normalization — is always positive, which makes

R

unique.

Worked Example

For

A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{pmatrix}

: Gram-Schmidt on the two columns gives

\mathbf{q}_1 = \frac{1}{\sqrt{2}}(1, 0, 1)^T

and

\mathbf{q}_2 = \frac{1}{\sqrt{6}}(1, 2, -1)^T

. Then

R = \begin{pmatrix} \sqrt{2} & 1/\sqrt{2} \\ 0 & 3/\sqrt{6} \end{pmatrix}

and

A = QR

.

Pitfall: Leaving the columns unnormalized

Putting

\mathbf{u}_1 = (1, 0, 1)^T

into

Q

without dividing by its length gives

Q^TQ \neq I

, and with

R_{11} = \sqrt{2}

the first column of

QR

becomes

(\sqrt{2}, 0, \sqrt{2})^T

instead of

\mathbf{a}_1

. With

\mathbf{q}_1 = \frac{1}{\sqrt{2}}(1, 0, 1)^T

it is

(1, 0, 1)^T

.

QR via Householder Reflections

A Householder reflection is an orthogonal matrix

H = I - 2\mathbf{v}\mathbf{v}^T/(\mathbf{v}^T\mathbf{v})

that reflects

\mathbb{R}^m

across the hyperplane perpendicular to

\mathbf{v}

. By choosing

\mathbf{v}

appropriately, a single Householder reflection zeros out all entries below the pivot in one column.

Applying Householder reflections sequentially — one per column — produces

H_n \cdots H_2 H_1 A = R

. Since each

H_i

is orthogonal,

Q = H_1 H_2 \cdots H_n

is orthogonal, giving

A = QR

.

Pitfall: Reversing the order of the reflections

Taking

Q = H_2H_1

from

H_2H_1A = R

puts

Q^T

in place of

Q

, and

QR

no longer returns

A

. Inverting the product reverses it,

(H_2H_1)^{-1} = H_1H_2

, so

A = H_1H_2R

.

Thin QR vs. Full QR

The thin (reduced) QR factorization has

Q_1

of size

m \times n

with orthonormal columns and

R_1

of size

n \times n

upper triangular:

A = Q_1 R_1

. This is the version produced by Gram-Schmidt and is sufficient for most applications.

The full QR factorization extends

Q_1

to a square

m \times m

orthogonal matrix

Q

by appending

m - n

columns forming an orthonormal basis for

\text{Col}(A)^\perp

. The factor

R

is extended to

m \times n

by appending

m - n

rows of zeros:

A = QR

.

The full version is needed when the orthogonal complement of the column space is required — for instance, when extracting a basis for the left null space. The thin version is more economical for system solving and least squares.

The two variants compare cleanly on the dimensions of their factors and what each one captures.

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Decompositions · QR

Thin QR and full QR

Both reconstruct $A$ exactly. The full form keeps $m - n$ extra columns of $Q$ that multiply against rows of zeros — useless for the product, and the only route to a basis for the left null space.

2forms

Type	Condition	Q	R	bases obtained	used for
The working form
Thin QR§ 4	$m \geq n$ , columns of $A$ independent	$m \times n$ , orthonormal columns	$n \times n$ upper triangular	a basis for $\operatorname{Col}(A)$	solving, least squares
The complete form
Full QR§ 4	same $A$ , $Q$ completed to a square	$m \times m$ , fully orthogonal	$m \times n$ — triangular block, then zeros	bases for $\operatorname{Col}(A)$ and its complement	left null space, theory

The extra columns are not redundant so much as unused. They complete

Q

into a genuine orthogonal matrix, which is what makes

QQ^{\mathsf{T}} = I

hold as well as

Q^{\mathsf{T}}Q = I

— and that second identity is what any argument about the left null space needs.

Thin QR and full QR·/linear-algebra/decompositions/qrLearn Math Class

The difference is

m - n

columns of

Q

paired against

m - n

rows of zeros in

R

, so they contribute nothing to the product and

A

comes back identically either way. What they do contribute is a basis for everything

A

cannot reach — the left null space — and the squareness that makes

Q

orthogonal in both directions rather than only one.

Existence and Uniqueness

THEOREM

Every

m \times n

matrix with

m \geq n

and linearly independent columns has a thin QR factorization. Every

m \times n

matrix (regardless of rank) has a full QR factorization.

The thin QR factorization with positive diagonal entries on

R

is unique. If negative diagonal entries are permitted, the factorization is not unique — signs can be redistributed between

Q

and

R

(multiplying a column of

Q

by

-1

and the corresponding row of

R

by

-1

preserves the product). The convention of positive diagonal entries on

R

resolves this ambiguity.

Solving Least Squares with QR

The normal equations

A^TA\hat{\mathbf{x}} = A^T\mathbf{b}

transform under

A = QR

. Since

A^TA = R^TQ^TQR = R^TR

and

A^T\mathbf{b} = R^TQ^T\mathbf{b}

, the normal equations become

R^TR\hat{\mathbf{x}} = R^TQ^T\mathbf{b}

. Canceling

R^T

(invertible because

R

has positive diagonal):

Solving Least Squares with QR

R\hat{\mathbf{x}} = Q^T\mathbf{b}

The right-hand side

Q^T\mathbf{b}

is computed by

n

dot products. The system

R\hat{\mathbf{x}} = Q^T\mathbf{b}

is upper triangular, solved by back substitution in

O(n^2)

operations.

The critical advantage over the normal equations is numerical.

RULE

Forming

A^TA

squares the condition number:

\kappa(A^TA) = \kappa(A)^2

.

If

A

has condition number

10^6

, the normal equations work with condition number

10^{12}

, losing

12

digits of accuracy in double precision. QR avoids this squaring and works with the original condition number

10^6

.

Least squares through QR needs one triangular solve and keeps the accuracy that forming

A^TA

throws away.

The QR Algorithm for Eigenvalues

The QR algorithm is the standard method for computing eigenvalues of general (non-symmetric) matrices. It proceeds iteratively, starting from

A_0 = A

and at each step factoring and reassembling the product in reversed order:

QR Algorithm for Eigenvalues

A_k = Q_k R_k, \qquad A_{k+1} = R_k Q_k

Qr algorithm for eigenvalues →

THEOREM

Under mild conditions,

A_k

converges to an upper triangular matrix with the eigenvalues on the diagonal.

Fact: Its inventor never saw it take over

John Francis published the QR algorithm in 1961, and Vera Kublanovskaya found it independently the same year. Francis left numerical analysis soon after and learned only in 2007 that his method had been named one of the ten most important algorithms of the 20th century.

The convergence is driven by the fact that

A_{k+1} = Q_k^T A_k Q_k

— each iteration is a similarity transformation that preserves the eigenvalues while driving the sub-diagonal entries toward zero.

With shifts (replacing

A_k

by

A_k - \sigma_k I

before factoring and adding

\sigma_k I

back), convergence accelerates dramatically — cubic convergence for symmetric matrices with the Wilkinson shift. The QR algorithm computes eigenvalues without ever forming the characteristic polynomial, avoiding the severe numerical instability of polynomial root-finding.

Reversing the factors is a similarity, so the QR algorithm can reshape a matrix freely without ever touching its eigenvalues.

Properties of the Factors

The orthonormality of

Q

's columns (

Q^TQ = I_n

) has several immediate consequences.

The matrix

QQ^T

is the projection matrix onto

\text{Col}(A)

. For any

\mathbf{b}

,

QQ^T\mathbf{b}

is the orthogonal projection of

\mathbf{b}

onto the column space.

Pitfall: Treating QQᵀ as the identity

Replacing

QQ^T

by

I

for a thin

Q

claims

QQ^T\mathbf{b} = \mathbf{b}

for every

\mathbf{b}

. For the worked example and

\mathbf{b} = (1, 1, 1)^T

, the product is

(4/3, 2/3, 2/3)^T

; only

Q^TQ = I_2

holds.

Gram-Schmidt vs. Householder

Classical Gram-Schmidt can lose orthogonality in floating-point arithmetic when the columns of

A

are nearly dependent. The computed

\mathbf{q}_i

's may fail to be perpendicular to machine precision, and the errors accumulate with each step.

Modified Gram-Schmidt improves stability by updating the remaining vectors in place after each projection subtraction, rather than using the original columns throughout. The mathematical result is identical in exact arithmetic, but the numerical behavior is significantly better.

Fact: At most

n

mirrors make any orthogonal matrix

Householder QR on an

n \times n

orthogonal matrix, with each diagonal entry kept positive, ends after

n - 1

reflections with an

R

that is triangular and orthogonal: the identity, except perhaps a

-1

in the last corner, which is one more reflection. So every orthogonal matrix is a product of at most

n

reflections.

Householder reflections provide the strongest stability guarantee. Each reflection zeros an entire column below the diagonal in a single, orthogonally-implemented step. The resulting QR factorization is backward stable — the gold standard in numerical linear algebra.

Givens rotations offer a third option, zeroing entries one at a time via plane rotations. They are preferred for sparse matrices, where surgically placed zeros can be introduced without disturbing the existing sparsity structure.

In practice, Householder is the default for dense matrices, Givens for sparse ones, and Gram-Schmidt (modified) for situations where the orthogonal factor

Q

is needed explicitly rather than implicitly.

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Before you read

Same result, different stability

All three methods produce the same QR factorization in exact arithmetic. In floating-point arithmetic they rank by how well the computed columns of

Q

stay orthogonal.

most stablenumerical stabilityleast stable

Householder reflectionsbackward stable

Modified Gram-Schmidtsignificantly better

Classical Gram-Schmidtcan lose orthogonality

Takeaway

In practice

Householder is the default for dense matrices, Givens rotations for sparse ones, and modified Gram-Schmidt when

Q

is needed explicitly.

QR methods ranked by numerical stability·/linear-algebra/decompositions/qrLearn Math Class

QR and Gram-Schmidt: The Connection

The Gram-Schmidt process and the QR decomposition are two descriptions of the same computation.

Gram-Schmidt takes the columns of

A

and produces orthonormal vectors

\mathbf{q}_1, \dots, \mathbf{q}_n

while recording the coefficients

R_{ij} = \mathbf{q}_i \cdot \mathbf{a}_j

along the way. Assembling these into matrices gives

A = QR

.

Conversely, given

A = QR

with positive diagonal entries on

R

, the columns of

Q

are exactly what Gram-Schmidt would produce, and

R

stores exactly the dot products Gram-Schmidt would compute. The factorization is the matrix-level summary of the vector-level algorithm.

This duality means every theorem about QR has an interpretation in terms of Gram-Schmidt, and vice versa. The QR decomposition is Gram-Schmidt made systematic, portable, and computable in a single matrix equation.

The leftover normalised to length one

What remained after the projections was a direction with an arbitrary length; dividing by that length makes it a unit vector and the next column of Q. Gram-Schmidt and QR are not two procedures but one, read as a geometric construction or as a factorisation. Step through the orthogonalisation on the Gram-Schmidt visualizer.

Seeing them as one procedure explains why improving the stability of Gram-Schmidt improves QR at the same time.

Summary: Four Routes to the Same Factorization

The QR factorization is unique (for linearly independent columns, with the positive-diagonal convention), but the algorithms that produce it are not. Classical Gram-Schmidt, modified Gram-Schmidt, Householder reflections, and Givens rotations all yield the same

A = QR

in exact arithmetic once signs are normalized so the diagonal of

R

is positive, but they differ sharply in numerical stability and in the kinds of matrices they handle most efficiently. The table below collects each method alongside how it operates, its stability behavior, and the setting in which it is the right choice.

The four methods divide cleanly by strategy. The first two build

Q

directly, orthogonalizing the columns of

A

one at a time; the second two leave

Q

implicit and instead apply orthogonal operations to

A

until what remains is triangular. That difference is what decides the stability, so the grouping below is also the ranking.

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Decompositions · QR

Four routes to the same factorization

All four produce $A = QR$ and none is a variant of the others in output — they differ only in how they get there and how much accuracy survives the trip. The split below is by numerical standing, which is the column that decides in practice.

4methods

Orthogonalization — build Q a column at a time2

1

Classical Gram–Schmidt§ 2

unstable — loses orthogonality

\mathbf{u}_j = \mathbf{a}_j - \sum_{i<j} \operatorname{proj}_{\mathbf{q}_i}(\mathbf{a}_j)

,

\mathbf{q}_j = \mathbf{u}_j / \|\mathbf{u}_j\|

Each column is orthogonalized against the original earlier columns. In exact arithmetic this is correct; in floating point the accumulated rounding means

Q^{\mathsf{T}}Q

drifts measurably from

I

, and badly when the columns are nearly dependent. Worth knowing as the definition, not as a method.

2

Modified Gram–Schmidt§ 9

better, still not backward stable

subtract each projection immediately

The same arithmetic reassociated: update the remaining columns in place after each subtraction rather than accumulating. Substantially more accurate than the classical form for no extra cost, which is why it is the version to use if Gram–Schmidt is used at all.

Triangularization — zero out A instead2

3

Householder reflections§ 3

backward stable — the default

H = I - 2\mathbf{v}\mathbf{v}^{\mathsf{T}} / \mathbf{v}^{\mathsf{T}}\mathbf{v}

One reflection per column zeros everything below the pivot at once.

Q

arrives as a product of reflections rather than being built directly, which is what keeps it orthogonal to machine precision. This is what a numerical library calls unless told otherwise.

4

Givens rotations§ 3

backward stable; local

a plane rotation per entry

Zeros one entry at a time instead of a whole column. Slower on a dense matrix, but each rotation touches only two rows — so a sparse matrix stays sparse, and the work can be parallelised. The reason to choose it is structure, not speed.

Two of these are used and two are taught. Gram–Schmidt makes the connection between orthogonalization and QR visible, which is why it appears in every course; Householder and Givens are what libraries actually call, because backward stability is the property that matters once the matrix is anything other than well behaved.

Four routes to the same factorization·/linear-algebra/decompositions/qrLearn Math Class

Why the second group wins is worth stating plainly. Building

Q

by orthogonalization means every column inherits the rounding error of the columns before it, so orthogonality degrades as the computation proceeds — and degrades fastest exactly when the columns are nearly dependent, which is when the factorization was most needed. Householder and Givens never construct

Q

from the data at all: each step is an orthogonal matrix by construction, and a product of orthogonal matrices is orthogonal whatever the arithmetic did.

In practice the choice is narrow. Householder is the default for dense matrices and is what least squares routines call. Givens is chosen when the matrix is sparse or when entries arrive one at a time, since a rotation acts on two rows and leaves the rest untouched. Modified Gram–Schmidt survives where the columns of

Q

are wanted explicitly and progressively — and classical Gram–Schmidt survives as an explanation.

Common Pitfalls

Three mistakes come up again and again when a QR factorization is computed by hand or used afterwards. Each is marked with a box where it happens in the sections above. Here each is worked through in full, with the wrong and the right working side by side.

Leaving the columns unnormalized

Gram-Schmidt first produces the orthogonal vectors

\mathbf{u}_j

, and these are put into

Q

as they are. The division by

\|\mathbf{u}_j\|

that turns them into unit vectors is skipped, while

R

is filled in with the lengths as usual.

Wrong working: for

A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{pmatrix}

, take

\mathbf{u}_1 = (1, 0, 1)^T

and

\mathbf{u}_2 = (1/2, 1, -1/2)^T

as the columns of

Q

, with

R = \begin{pmatrix} \sqrt{2} & 1/\sqrt{2} \\ 0 & 3/\sqrt{6} \end{pmatrix}

. Then

Q^TQ = \begin{pmatrix} 2 & 0 \\ 0 & 3/2 \end{pmatrix} \neq I, \qquad \text{first column of } QR = \sqrt{2}\,\mathbf{u}_1 = (\sqrt{2}, 0, \sqrt{2})^T \neq \mathbf{a}_1

Right working: divide each

\mathbf{u}_j

by its length.

\mathbf{q}_1 = \frac{\mathbf{u}_1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(1, 0, 1)^T, \qquad \mathbf{q}_2 = \frac{\mathbf{u}_2}{\sqrt{3/2}} = \frac{1}{\sqrt{6}}(1, 2, -1)^T

Now

Q^TQ = I_2

and

QR = A

. The first column of

QR

is

\sqrt{2}\,\mathbf{q}_1 = (1, 0, 1)^T

, and the second is

\frac{1}{\sqrt{2}}\mathbf{q}_1 + \frac{3}{\sqrt{6}}\mathbf{q}_2 = (\frac{1}{2}, 0, \frac{1}{2})^T + (\frac{1}{2}, 1, -\frac{1}{2})^T = (1, 1, 0)^T

.

Why it happens: the length

\|\mathbf{u}_j\|

is already recorded in

R

as the diagonal entry, so it feels as if it has been used. It is needed twice: once as

R_{jj}

, and once as the divisor that makes

\mathbf{q}_j

a unit vector.

How to check: compute

Q^TQ

. Its diagonal entries are the squared lengths of the columns of

Q

, and every one of them must be

1

.

Reversing the order of the reflections

Householder reflections are applied to

A

one after another, giving

H_n \cdots H_2 H_1 A = R

. The product on the left is then taken as

Q

in the same order, although

Q

has to undo it.

Wrong working: take

A = \begin{pmatrix} 2 & 5 \\ 1 & 0 \\ 2 & -2 \end{pmatrix}

. The reflections

H_1 = \frac{1}{3}\begin{pmatrix} 2 & 1 & 2 \\ 1 & 2 & -2 \\ 2 & -2 & -1 \end{pmatrix}, \qquad H_2 = \frac{1}{5}\begin{pmatrix} 5 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & -3 \end{pmatrix}

give

H_1A = \begin{pmatrix} 3 & 2 \\ 0 & 3 \\ 0 & 4 \end{pmatrix}

and

H_2H_1A = R = \begin{pmatrix} 3 & 2 \\ 0 & 5 \\ 0 & 0 \end{pmatrix}

. Taking

Q = H_2H_1

:

H_2H_1 = \frac{1}{15}\begin{pmatrix} 10 & 5 & 10 \\ 11 & -2 & -10 \\ -2 & 14 & -5 \end{pmatrix}, \qquad H_2H_1R = \begin{pmatrix} 2 & 3 \\ 11/5 & 4/5 \\ -2/5 & 22/5 \end{pmatrix} \neq A

Right working: move the product to the other side. Each

H_i

is its own inverse, so

A = (H_2H_1)^{-1}R = H_1^{-1}H_2^{-1}R = H_1H_2R

, and

Q = H_1H_2 = \frac{1}{15}\begin{pmatrix} 10 & 11 & -2 \\ 5 & -2 & 14 \\ 10 & -10 & -5 \end{pmatrix}, \qquad QR = \begin{pmatrix} 2 & 5 \\ 1 & 0 \\ 2 & -2 \end{pmatrix} = A

The wrong product

H_2H_1

is exactly

Q^T

.

Why it happens:

H_2H_1

is the matrix that appears on the page, and it is orthogonal, so it looks like a valid

Q

. The inversion that reverses the order is a single step that is easy to skip.

How to check: the first column of

Q

must be the first column of

A

divided by

R_{11}

. Here that is

\frac{1}{3}(2, 1, 2)^T

, which is the first column of

H_1H_2

and not of

H_2H_1

.

Treating QQᵀ as the identity

For a thin

Q

with

m > n

, the identity

Q^TQ = I_n

is turned around and

QQ^T

is replaced by

I_m

, as if

Q

were square.

Wrong working: with

Q

from the worked example and

\mathbf{b} = (1, 1, 1)^T

, the least-squares solution

\hat{\mathbf{x}} = R^{-1}Q^T\mathbf{b}

gives

A\hat{\mathbf{x}} = QRR^{-1}Q^T\mathbf{b} = QQ^T\mathbf{b}

, and replacing

QQ^T

by

I

makes this

\mathbf{b}

. The fit looks exact, with zero residual.

Right working:

QQ^T\mathbf{b}

is the projection of

\mathbf{b}

onto

\text{Col}(A)

.

QQ^T\mathbf{b} = (\mathbf{q}_1 \cdot \mathbf{b})\,\mathbf{q}_1 + (\mathbf{q}_2 \cdot \mathbf{b})\,\mathbf{q}_2 = (1, 0, 1)^T + (\tfrac{1}{3}, \tfrac{2}{3}, -\tfrac{1}{3})^T = (\tfrac{4}{3}, \tfrac{2}{3}, \tfrac{2}{3})^T

Solving

R\hat{\mathbf{x}} = Q^T\mathbf{b} = (\sqrt{2}, 2/\sqrt{6})^T

gives

\hat{\mathbf{x}} = (2/3, 2/3)^T

, and

A\hat{\mathbf{x}} = (4/3, 2/3, 2/3)^T

. The residual is

\mathbf{b} - A\hat{\mathbf{x}} = (-1/3, 1/3, 1/3)^T

, not zero.

Why it happens: for a square orthogonal matrix both products equal

I

, and the two identities are learned together. For a thin

Q

only the product of size

n \times n

is the identity.

How to check: compare sizes and rank.

QQ^T

is

m \times m

but has rank only

n

, so for

m > n

it cannot be

I_m

. The residual

\mathbf{b} - QQ^T\mathbf{b}

is zero only when

\mathbf{b}

already lies in

\text{Col}(A)

.

QR Decomposition FAQ

Is a thin $Q$ an orthogonal matrix?

+

No, that name requires a square matrix, and a thin

Q

has none of the properties it implies, including an inverse. The practical form of the error is cancelling

QQ^{T}

inside a longer expression as though it were the identity. For a thin

Q

only

Q^{T}Q = I

holds, while

QQ^{T}

is a projection matrix.Read more →

Is the QR factorisation unique?

+

Only once a sign convention is fixed, usually requiring the diagonal of

R

to be positive. Gram-Schmidt satisfies that automatically since each diagonal entry is a norm, but Householder reflections do not, and numerical libraries rarely enforce it. Two correct factorisations of one matrix can therefore differ by a column of sign flips.Read more →

In the QR algorithm, is $R_kQ_k$ a second factorisation?

+

No, it is a deliberate reversal of the factors just computed, and reversing them changes the matrix rather than undoing anything. Repeating that swap is exactly what drives the sequence toward upper triangular form with the eigenvalues along the diagonal. The subscript

k

counts iterations rather than rows or columns.Read more →

QR Decompositions

Orthogonal Times Triangular

What QR Decomposition Is

QR Notation

QR Notation

QR via Gram-Schmidt

QR via Householder Reflections

Thin QR vs. Full QR

Thin QR and full QR

Existence and Uniqueness

Solving Least Squares with QR

The QR Algorithm for Eigenvalues

Properties of the Factors

Gram-Schmidt vs. Householder

QR and Gram-Schmidt: The Connection

Summary: Four Routes to the Same Factorization

Four routes to the same factorization

Common Pitfalls

QR Decomposition FAQ

Is a thin $Q$ an orthogonal matrix?

Is the QR factorisation unique?

In the QR algorithm, is $R_kQ_k$ a second factorisation?

QR Decompositions

Orthogonal Times Triangular

What QR Decomposition Is

QR Notation

QR Notation

QR via Gram-Schmidt

QR via Householder Reflections

Thin QR vs. Full QR

Thin QR and full QR

Existence and Uniqueness

Solving Least Squares with QR

The QR Algorithm for Eigenvalues

Properties of the Factors

Gram-Schmidt vs. Householder

QR and Gram-Schmidt: The Connection

Summary: Four Routes to the Same Factorization

Four routes to the same factorization

Common Pitfalls

QR Decomposition FAQ

Is a thin QQQ an orthogonal matrix?

Is the QR factorisation unique?

In the QR algorithm, is RkQkR_kQ_kRk​Qk​ a second factorisation?

Is a thin $Q$ an orthogonal matrix?

In the QR algorithm, is $R_kQ_k$ a second factorisation?