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Positive-diagonal QR factors give the exact finite-time tangent-volume change

Statement

Let $A_{1}, \dots, A_{N} \in R^{n \times n}$ be invertible, with $n, N \geq 1$ , and let $Q_{0}$ be orthogonal. For $1 \leq k \leq N$ , form the unique positive-diagonal QR factorization $A_{k} Q_{k - 1} = Q_{k} R_{k},$ where $Q_{k}$ is orthogonal and $R_{k}$ is upper triangular with positive diagonal $r_{k, i i}$ . Put $P_{N} = A_{N} \dots A_{1}$ and define the exact finite-time log-volume change by $\log ∣ \det P_{N} ∣$ . Then $\log ∣ \det P_{N} ∣ = \sum_{k = 1}^{N} \sum_{i = 1}^{n} \log r_{k, i i} .$ No infinite-time Lyapunov exponent or shadowing conclusion follows.

For a numerical log record, define $σ_{\min} (A) : = \inf_{∥ x ∥_{2} = 1} ∥ A x ∥_{2}$ . If $σ_{\min} (A_{k}) \geq σ > 0$ , each exact $r_{k, i i} \geq σ$ . If positive computed diagonals ${\tilde{r}}_{k, i i}$ satisfy $∣ {\tilde{r}}_{k, i i} - r_{k, i i} ∣ \leq ε_{k, i} \leq σ / 2$ , and each computed logarithm ${\hat{l}}_{k, i}$ satisfies $∣ {\hat{l}}_{k, i} - \log {\tilde{r}}_{k, i i} ∣ \leq δ_{k, i}$ , then $∣ \sum_{k, i} {\hat{l}}_{k, i} - \log ∣ \det P_{N} ∣ ∣ \leq \sum_{k, i} (\frac{2 ε_{k, i}}{σ} + δ_{k, i}) .$ If any factor loses rank, this log-volume branch returns failure.

Facts & Assumptions

Given: The finite list of invertible tangent matrices, an orthogonal $Q_{0}$ , and the stated factorizations. The numerical bound additionally assumes the displayed certified singular-value and log-evaluation errors.

[F1]

Every invertible real square matrix has a unique factorization $A = Q R$ with $Q$ orthogonal and $R$ upper triangular with positive diagonal (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).

[F2]

Determinants multiply under finite square matrix products (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[F3]

The determinant of an upper triangular matrix is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[F4]

An orthogonal matrix has determinant modulus one (Orthogonal and unitary operators form groups, and their determinants have modulus one).

[F5]

The logarithm difference is the integral of $1 / t$ between positive endpoints (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[A1]

For the numerical bound, $σ_{\min} (A_{k}) \geq σ > 0$ , $∣ {\tilde{r}}_{k, i i} - r_{k, i i} ∣ \leq ε_{k, i} \leq σ / 2$ , and the computed logarithm errors are bounded by $δ_{k, i}$ .

Proof

1.1givenalgebrainduction

Induction on $k$ using $A_{k} Q_{k - 1} = Q_{k} R_{k}$ gives $P_{k} Q_{0} = Q_{k} R_{k} R_{k - 1} \dots R_{1}, P_{k} : = A_{k} \dots A_{1} .$ For $k = 1$ this is the defining factorization; multiplying the identity at $k$ on the left by $A_{k + 1}$ and substituting $A_{k + 1} Q_{k} = Q_{k + 1} R_{k + 1}$ proves the next case.

1.2F1givenalgebra

Orthogonal factors preserve Euclidean norms, so $σ_{\min} (R_{k}) = σ_{\min} (Q_{k} R_{k}) = σ_{\min} (A_{k} Q_{k - 1}) = σ_{\min} (A_{k}) \geq σ$ . For a fixed $i$ , the leading $(i - 1) \times (i - 1)$ block of $R_{k}$ is invertible. Choose $x \in R^{i - 1}$ solving $R_{1 : i - 1, 1 : i - 1} x = - R_{1 : i - 1, i}$ and set $z = (x, 1, 0, \dots, 0)^{T}$ . Upper triangularity gives $R_{k} z = r_{k, i i} e_{i}$ and $∥ z ∥_{2} \geq 1$ , so $σ_{\min} (R_{k}) \leq ∥ R_{k} z ∥_{2} / ∥ z ∥_{2} \leq r_{k, i i}$ . For $i = 1$ , take $z = e_{1}$ . Hence $r_{k, i i} \geq σ$ .

2.1F2F3F4step 1.1algebra

Take determinants in step 1.1. By [F2], $\det (P_{N}) \det (Q_{0}) = \det (Q_{N}) \prod_{k = 1}^{N} \det (R_{k})$ . By [F3] each $\det (R_{k}) = \prod_{i} r_{k, i i} > 0$ , and by [F4] $∣ \det (Q_{0}) ∣ = ∣ \det (Q_{N}) ∣ = 1$ . Thus $∣ \det (P_{N}) ∣ = \prod_{k, i} r_{k, i i}$ . Taking logarithms of positive numbers proves the exact identity.

3.1A1F5step 1.2algebra∎

Both $r_{k, i i}$ and ${\tilde{r}}_{k, i i}$ are at least $σ / 2$ . By [F5], $∣ \log {\tilde{r}}_{k, i i} - \log r_{k, i i} ∣ \leq (2 / σ) ∣ {\tilde{r}}_{k, i i} - r_{k, i i} ∣$ . Adding the computation error $δ_{k, i}$ and summing the scalar bounds by the triangle inequality proves the stated numerical log bound. The lower singular-value bound is essential to keep the log derivative bounded; if any factor is singular its determinant is zero and its log-volume is undefined.

Depends on

Used by

A finite QR record recovers the tangent-volume change Example

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Predrag Cvitanović et al., ChaosBook, Chapter 6: Lyapunov exponents, §6.2 (standard reference, not scraped)
Sheldon Axler, Linear Algebra Done Right, 4th ed., §7D, result 7.58 (standard reference, not scraped)