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A stable-spectrum nonnormal exponential has a large transient norm

Example

Let $A = (\begin{array}{cc} - 1 & 10 \\ 0 & - 1 \end{array}) .$ Its eigenvalues have negative real part, but at time $t = 1$ the row-sum norm of $e^{A}$ is greater than one. Here the row-sum norm is $∥ M ∥_{r o w} = \max_{i} \sum_{j} ∣ M_{i j} ∣$ .

Facts & Assumptions

Given: The displayed matrix $A$ .

[F1]

The complex exponential is its power-series exponential and its primary matrix value is given by the Jordan-block formula (The complex exponential by its power series, Jordan, contour, similarity, and normal-functional-calculus formulas agree).

[F2]

The exponential series is absolutely convergent for a bounded matrix operator (The exponential series of a bounded operator).

Verification

1.1givenF1algebra

Write $A = - I + 10 N$ with $N = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ and $N^{2} = 0$ . For $S = diag (10, 1)$ , $A = S (- I + N) S^{- 1}$ . The Jordan formula and similarity invariance in [F1] therefore give $e^{A} = e^{- 1} (I + 10 N) = e^{- 1} (\begin{array}{cc} 1 & 10 \\ 0 & 1 \end{array})$ . Direct multiplication gives $A^{*} A = (\begin{array}{cc} 1 & - 10 \\ - 10 & 101 \end{array})$ and $A A^{*} = (\begin{array}{cc} 101 & - 10 \\ - 10 & 1 \end{array})$ , so $A$ is nonnormal.

2.1F2step 1.1algebra∎

The row-sum norm is $11 / e$ , while the spectral radius of $e^{A}$ is $e^{- 1} < 1$ . From [F2], $e = \sum_{k \geq 0} 1 / k! < 3$ (for $k \geq 1$ , $k! \geq 2^{k - 1}$ , with strict inequality for some terms). Thus $11 / e > 1$ , displaying transient norm growth despite the stable eigenvalues.

Depends on

Used by

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Sources

Awad H. Al-Mohy and Nicholas J. Higham, A New Scaling and Squaring Algorithm for the Matrix Exponential (standard reference, not scraped)
Jan Zur, Matrix Functions, TU Berlin full lecture notes (standard reference, not scraped)