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Backward Euler scalar A-stability does not imply matrix contractivity

Statement refuted

For every real matrix $A$ whose characteristic polynomial has all roots in the closed left half-plane, the backward-Euler matrix step $(I - A)^{- 1}$ is contractive in the Euclidean norm.

Facts & Assumptions

Given: The real matrix $A = (\begin{array}{cc} - 1 & 16 \\ 0 & - 1 \end{array})$ and a unit backward-Euler step $h = 1$ .

[F1]

Matrix-step contractivity and the logarithmic norm in a specified inner product: Euclidean contractivity requires the induced norm of the matrix step to be at most one; $μ_{2} (A) = \sup_{x \neq 0} x^{T} A x / ∥ x ∥_{2}^{2}$ .

[F2]

Backward Euler is A-stable and L-stable: the backward-Euler scalar stability function is $R_{B E} (z) = 1 / (1 - z)$ and is A-stable.

Counterexample

1.1givenF2algebra

The characteristic polynomial of $A$ is $(ν + 1)^{2}$ , so its roots are both $- 1$ and the scalar backward-Euler factor is $R_{B E} (- 1) = 1 / 2$ . Also $A A^{T} \neq A^{T} A$ , so $A$ is nonnormal.

1.2givenF1algebra

Direct inversion gives $(I - A)^{- 1} = (\begin{array}{cc} 1 / 2 & 4 \\ 0 & 1 / 2 \end{array})$ . For $e_{2} = (0, 1)^{T}$ , this step gives $(4, 1 / 2)^{T}$ , whose Euclidean norm is $\sqrt{65 / 4} > 1 = ∥ e_{2} ∥_{2}$ . Thus the matrix step is not contractive.

2.1F1step 1.1algebra∎

For $x = (x_{1}, x_{2})^{T}$ , $7 ∥ x ∥_{2}^{2} - x^{T} A x = 8 (x_{1} - x_{2})^{2} \geq 0$ , with equality at $x_{1} = x_{2} \neq 0$ . Hence $μ_{2} (A) = 7 > 0$ , which records the nonnormal growth hidden by the negative scalar test value. The claimed implication is false.

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Sources

MIT OpenCourseWare, 18.330 Introduction to Numerical Analysis, Chapter 5, §5.2.2: Stability (standard reference, not scraped)
Ernst Hairer and Gerhard Wanner, Introduction à l’Analyse Numérique, Chapter III, §III.9 (standard reference, not scraped)