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Every invertible real or complex square matrix has a unique factorisation $A = Q R$ with $Q$ orthogonal or unitary and $R$ upper triangular with positive real diagonal

Statement

Every invertible matrix $A \in M_{n} (F)$ , where $F = R$ or $C$ , has a unique factorisation

$A = Q R,$

where $Q^{*} Q = I$ and $R$ is upper triangular with positive real diagonal entries. Thus $Q$ is orthogonal over $R$ and unitary over $C$ . The assertion includes the unique $0 \times 0$ factorisation.

Facts & Assumptions

Given: An invertible $n \times n$ matrix $A$ over $R$ or $C$ .

[L1]

The columns of an invertible square matrix form a basis of the coordinate space (A square matrix is invertible exactly when its multiplication map is a linear isomorphism; matrices preserve inverses of linear isomorphisms).

[L2]

Gram–Schmidt produces an orthonormal list with the same successive column spans and positive normalising factors (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).

[L3]

Matrix columns are the coordinate columns of the represented map on the standard basis (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[L4]

A square operator is invertible exactly when its determinant is nonzero (A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).

[L5]

The conjugate transpose is obtained by entrywise conjugation followed by transposition (The transpose $A^{T}$ of a matrix, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

For $n = 0$ , take the unique empty matrices $Q$ and $R$ . Now suppose $n > 0$ . By [L1], the columns $(a_{0}, \dots, a_{n - 1})$ of $A$ are independent. Apply [L2] to obtain an orthonormal basis $(q_{0}, \dots, q_{n - 1})$ with the same successive spans.

L1L2

1.2

Suppose $A = Q_{1} R_{1} = Q_{2} R_{2}$ are two such factorisations. The positive diagonal makes both $R_{i}$ invertible, so $U : = Q_{2}^{*} Q_{1} = R_{2} R_{1}^{- 1}$ is both unitary and upper triangular, with positive real diagonal.

L4L5algebra

2.1

Let $Q$ have columns $q_{j}$ and set $R_{i j} = ⟨ a_{j}, q_{i} ⟩$ . The successive-span property makes $R_{i j} = 0$ for $i > j$ , and the Gram–Schmidt normalisation gives $R_{j j} > 0$ . Expanding each $a_{j}$ in the orthonormal basis and using [L3] gives $A = Q R$ . Orthonormality gives $Q^{*} Q = I$ by [L5].

step 1.1L2L3L5

2.2

The first column of an upper-triangular unitary matrix has only its first entry nonzero; unit length and positive diagonal make that entry $1$ . Orthogonality with the remaining columns makes their first entries zero. Induction on the trailing principal block gives $U = I$ .

step 1.2induction

3.1

Hence $Q_{1} = Q_{2}$ and then $R_{1} = R_{2}$ . Steps 1.1 and 2.1 give existence, while steps 1.2 and 2.2 give uniqueness in every dimension.

step 1.1step 1.2step 2.1step 2.2∎

Depends on

Used by

A worked QR factorisation of an invertible real matrix Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 89 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., result 7.58 (standard reference, not scraped)

Every invertible real or complex square matrix has a unique factorisation A=QRA=QR with QQ orthogonal or unitary and RR upper triangular with positive real diagonal