ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generated

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A finite QR record recovers the tangent-volume change

Example

Take $Q_{0} = I_{2}$ , $A_{1} = diag (2, 1)$ , and $A_{2} = diag (1 / 3, 4)$ . Both positive-diagonal QR factorizations keep $Q_{k} = I_{2}$ and have $R_{k} = A_{k}$ .

Facts & Assumptions

Given: The two invertible diagonal tangent matrices and the finite-time QR identity.

[F1]

The positive-diagonal QR factors give the exact finite-time log-volume change as the sum of their log diagonals (Positive-diagonal QR factors give the exact finite-time tangent-volume change).

Verification

1.1givenalgebra

The product is $P_{2} = A_{2} A_{1} = diag (2 / 3, 4)$ , so $\log ∣ \det P_{2} ∣ = \log (8 / 3)$ . The positive-diagonal QR factors have diagonal entries $(2, 1)$ and $(1 / 3, 4)$ .

2.1F1step 1.1algebra∎

Their accumulated log sum is $\log 2 + \log 1 + \log (1 / 3) + \log 4 = \log (8 / 3)$ , which matches [F1]. The calculation is finite-time and asserts no limit.

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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)