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Computed dichotomy constants for a Jordan block and for a diagonal sink-source

Example

Use zero-based coordinates and the standard vectors $e_{0} = (1, 0)$ and $e_{1} = (0, 1)$ of $R^{2}$ .

A non-diagonalisable Jordan block. For

$A = (\begin{array}{cc} - 1 & 1 \\ 0 & - 1 \end{array})$

the exponential series gives $e^{t A} = e^{- t} (\begin{array}{cc} 1 & t \\ 0 & 1 \end{array})$ ; the only eigenvalue is $- 1$ , so the dichotomy theorem has $E_{s} = R^{2}$ , $E_{u} = {0}$ , and the constants $C = 1$ , $c = 1 / 2$ are admissible: for every $v \in R^{2}$ and $t \geq 0$ ,

$∥ e^{t A} v ∥_{2} \leq e^{- t / 2} ∥ v ∥_{2} .$

A diagonal sink-source. For $B = diag (1 / 2, - 3 / 2)$ the vectors $e_{0}, e_{1}$ are eigenvectors with eigenvalues $1 / 2$ and $- 3 / 2$ , which have nonzero real part, and the dichotomy theorem gives $E_{u} = R e_{0}$ , $E_{s} = R e_{1}$ with admissible constants $C = 1$ , $c = 1 / 4$ : for $t \geq 0$ one has $∥ e^{t B} e_{1} ∥_{2} = e^{- 3 t / 2} \leq e^{- t / 4}$ and $∥ e^{t B} e_{0} ∥_{2} = e^{t / 2} \geq e^{t / 4}$ , and for $t \leq 0$ one has $∥ e^{t B} e_{0} ∥_{2} = e^{t / 2} \leq e^{t / 4}$ .

Facts & Assumptions

Given: The matrices $A = (\begin{array}{cc} - 1 & 1 \\ 0 & - 1 \end{array})$ and $B = diag (1 / 2, - 3 / 2)$ , and the Euclidean norm on $R^{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L1]

Euclidean $R^{n}$ is complete ( $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ ), and a finite matrix is bounded: Cauchy-Schwarz row by row gives $∥ A v ∥_{2}^{2} \leq (\sum_{i, j < n} A_{i j}^{2}) ∥ v ∥_{2}^{2}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation). For $A \in M_{n} (R)$ the exponential series defines $E (t) = e^{t A}$ with $E (0) = I_{n}$ and $\frac{d}{d t} E (t) = A E (t) = E (t) A$ for all real $t$ (The exponential series of a bounded operator, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[L2]

For a fixed matrix $C \in M_{n} (R)$ the initial value problem $Y^{'} = C Y$ , $Y (0) = I_{n}$ has a unique solution on every compact interval containing $0$ , so a differentiable matrix curve satisfying these two conditions on $R$ equals $e^{t C}$ : apply uniqueness on $[- R, R]$ for arbitrary $R > 0$ (Linear matrix ODEs have unique global solutions on a fixed interval).

[L3]

The dichotomy theorem gives $T$ -invariant subspaces $E_{s}, E_{u}$ with $R^{n} = E_{s} \oplus E_{u}$ and constants $C \geq 1$ , $c > 0$ such that clauses 1-3 of the theorem hold, and $E_{s} = {v : e^{t T} v \to 0 (t \to + \infty)}$ , $E_{u} = {v : e^{t T} v \to 0 (t \to - \infty)}$ , whenever no eigenvalue of its complexification $T_{C} : C \otimes_{R} R^{n} \to C \otimes_{R} R^{n}$ has real part $0$ (Exponential dichotomy for a hyperbolic endomorphism: invariant stable and unstable subspaces with exponential estimates); $T_{C}$ is the complex-linear extension of $T$ (Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding, Complexification of a real-linear map), represented by the same matrix on $C^{n}$ in the standard coordinate identification (the dichotomy theorem, Fact F1); $λ \in C$ is its eigenvalue when $T_{C} v = λ v$ for some $0 \neq v \in C^{n}$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

[L4]

Scalar exponential: $e^{x + y} = e^{x} e^{y}$ , $e^{- x} = 1 / e^{x} > 0$ and $1 + s \leq e^{s}$ for all real $x, y, s$ ; $e^{x} \to + \infty$ as $x \to + \infty$ and $e^{x} \to 0$ as $x \to - \infty$ ; $(e^{x})^{'} = e^{x}$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ , The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ , The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[L5]

A continuous real function on an interval that is differentiable in its interior and has nonpositive derivative is nonincreasing (On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed).

Verification

technique · direct; solve the linear matrix initial value problems explicitly and check the dichotomy estimates

1.1givenL1L2L4algebra

Write $A = - I_{2} + N$ with $N = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ , so that $N^{2} = 0$ and $N$ commutes with $I_{2}$ . Define $F (t) : = e^{- t} (I_{2} + t N) = e^{- t} (\begin{array}{cc} 1 & t \\ 0 & 1 \end{array})$ ; differentiating entrywise with the product rule and $(e^{- t})^{'} = - e^{- t}$ [L4] gives $F^{'} (t) = - e^{- t} (I_{2} + t N) + e^{- t} N = e^{- t} (- I_{2} + N - t N)$ and $A F (t) = (- I_{2} + N) e^{- t} (I_{2} + t N) = e^{- t} (- I_{2} + N - t N + t N^{2}) = e^{- t} (- I_{2} + N - t N)$ ; since $F (0) = I_{2}$ , [L2] gives $e^{t A} = F (t)$ for all $t$ .

1.2givenL2L3L4algebra

For $B = diag (1 / 2, - 3 / 2)$ define $G (t) : = diag (e^{t / 2}, e^{- 3 t / 2})$ ; differentiating entrywise gives $G^{'} (t) = diag (\frac{1}{2} e^{t / 2}, - \frac{3}{2} e^{- 3 t / 2}) = B G (t)$ and $G (0) = I_{2}$ , so [L2] gives $e^{t B} = G (t)$ ; in particular $e^{t B} e_{0} = e^{t / 2} e_{0}$ and $e^{t B} e_{1} = e^{- 3 t / 2} e_{1}$ . No eigenvalue of $B$ is purely imaginary: if $B_{C} v = λ v$ with $λ \in C$ and $0 \neq v = (v_{0}, v_{1}) \in C^{2}$ , the two coordinates give $\frac{1}{2} v_{0} = λ v_{0}$ and $- \frac{3}{2} v_{1} = λ v_{1}$ , so $v_{0} \neq 0$ forces $λ = 1 / 2$ while $v_{1} \neq 0$ forces $λ = - 3 / 2$ . Hence [L3] applies to $B$ .

2.1step 1.1L3algebra

The only eigenvalue of $A$ is $- 1$ : if $A_{C} v = λ v$ with $λ \in C$ and $0 \neq v = (v_{0}, v_{1}) \in C^{2}$ , the second coordinate gives $- v_{1} = λ v_{1}$ and the first gives $- v_{0} + v_{1} = λ v_{0}$ . If $v_{1} = 0$ then $v_{0} \neq 0$ and $λ = - 1$ ; if $v_{1} \neq 0$ then $λ = - 1$ from the second coordinate. So no eigenvalue of $A$ has real part $0$ , and $[L 3]$ applies to $A$ .

2.2step 1.1L3L4L5algebra

Norm estimate for $A$ : put $w (t) : = e^{t A} v$ , so $w^{'} = A w$ by step 1.1. For $f (t) : = ∥ w (t) ∥_{2}^{2} = w_{0} (t)^{2} + w_{1} (t)^{2}$ the product rule gives $f^{'} (t) = 2 ⟨ A w, w ⟩_{1} = - 2 w_{0}^{2} + 2 w_{0} w_{1} - 2 w_{1}^{2} \leq - w_{0}^{2} - w_{1}^{2} = - f (t),$ because $2 w_{0} w_{1} \leq w_{0}^{2} + w_{1}^{2}$ , equivalently $(w_{0} - w_{1})^{2} \geq 0$ . Hence $(e^{t} f (t))^{'} = e^{t} (f + f^{'}) \leq 0$ , and the derivative sign criterion [L5] gives $e^{t} f (t) \leq f (0) = ∥ v ∥_{2}^{2}$ for $t \geq 0$ . Taking nonnegative square roots yields $∥ e^{t A} v ∥_{2} \leq e^{- t / 2} ∥ v ∥_{2}$ for all $v$ and $t \geq 0$ , including $v = 0$ . Thus $C = 1$ , $c = 1 / 2$ are admissible constants for $A$ in the sense of [L3].

3.1step 2.12.2L3L4

Identification of the subspaces for $A$ : the estimate of step 2.2 holds for every $v \in R^{2}$ , so $e^{t A} v \to 0$ as $t \to + \infty$ for every $v$ (by $e^{- t / 2} \to 0$ from [L4]); by the characterisation in [L3] the stable subspace is $E_{s} = R^{2}$ , and since $R^{2} = E_{s} \oplus E_{u}$ we get $E_{u} = {0}$ .

4.1step 2.11.2L3L4algebra∎

Subspaces and constants for $B$ : for $v = (v_{0}, v_{1})$ step 1.2 gives $e^{t B} v = (e^{t / 2} v_{0}, e^{- 3 t / 2} v_{1})$ ; from the limits in [L4], as $t \to + \infty$ this tends to $0$ exactly when $v_{0} = 0$ , and as $t \to - \infty$ exactly when $v_{1} = 0$ . By [L3], $E_{s} = R e_{1}$ and $E_{u} = R e_{0}$ . With $c = 1 / 4$ the estimates hold: for $t \geq 0$ , $e^{- 3 t / 2} \leq e^{- t / 4}$ and $e^{t / 2} \geq e^{t / 4}$ ; for $t \leq 0$ , $e^{t / 2} \leq e^{t / 4}$ . Each follows from $e^{a} \leq e^{b}$ whenever $a \leq b$ , which is a consequence of $e^{b} = e^{a} e^{b - a}$ and $e^{b - a} \geq 1$ for $b - a \geq 0$ [L4]. Hence $C = 1$ , $c = 1 / 4$ are admissible constants for $B$ . This proves the computed constants.

Remarks

The first matrix shows that the dichotomy does not require diagonalisability: the explicit shear has a polynomial factor in its entries, while the symmetric-part estimate in step 2.2 proves the Euclidean bound with $C = 1$ , $c = 1 / 2$ . The constant $C$ is not claimed to be $1$ for every admissible rate: the computation is a finite check for the displayed rates, not a general bound.
In both examples the admissible rates are strictly below the spectral gap, and the conclusion of [L3] is realized with $C = 1$ ; no claim is made about optimal constants or about the converse direction of the theorem.

Computed dichotomy constants for a Jordan block and for a diagonal sink-source

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

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