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Sylvester and lyapunov matrix equations — Examples

1 · Prerequisites

2 · Summary

The examples compute the separation and solution of a Sylvester equation with nearly colliding eigenvalue sums, the positive matrix solving a stable two-dimensional Lyapunov covariance balance, and a one-step low-rank ADI approximation whose residual factor is exactly zero. The covariance example checks the algebraic balance and positivity; it makes no separate stochastic-realization claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

A stable linear noise covariance balance has a positive solution

Example

For $A = diag (- 1, - 2)$ and noise-factor column $B = (1, 1)^{T}$ , the stable covariance-balance equation is $A X + X A^{T} + B B^{T} = 0$ . Its unique positive solution is

X = (\begin{array}{cc} 1 / 2 & 1 / 3 \\ 1 / 3 & 1 / 4 \end{array}) .

This verifies the algebraic covariance balance and its positive solution; it does not assert a separate stochastic realization or stationary-law theorem.

Facts & Assumptions

Given: The matrices in the Example.

[F1]

Hurwitz matrices and the continuous Lyapunov equation: $A$ is Hurwitz when all its eigenvalues have negative real part, and the balance is $A X + X A^{T} + Q = 0$ .

[F2]

The Hurwitz Lyapunov integral is the unique solution and has the stated positivity: For Hurwitz $A$ , the Lyapunov equation has a unique solution, and the integral solution is positive definite under the stated conditions.

Verification

1.1givenF1algebra

The eigenvalues of $A$ are $- 1$ and $- 2$ , so $A$ is Hurwitz by [F1]. Direct multiplication gives $B B^{T} = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ and $A X + X A^{T} = (\begin{array}{cc} - 1 & - 1 \\ - 1 & - 1 \end{array}) = - B B^{T}$ . Thus the displayed $X$ solves the balance.

2.1step 1.1algebra

For real $(u, v)$ , $(\begin{array}{cc} u & v \end{array}) X (\begin{array}{c} u \\ v \end{array}) = \frac{1}{2} (u + \frac{2}{3} v)^{2} + \frac{1}{36} v^{2}$ , which is positive unless $u = v = 0$ . Hence $X ≻ 0$ .

3.1F1F2step 1.1step 2.1∎

Since $A$ is Hurwitz, [F2] gives uniqueness of the solution, so this positive matrix is the unique stable Lyapunov covariance-balance solution.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Nearly colliding eigenvalue sums make a Sylvester solve sensitive

Example

Let $ε = 10^{- 6}$ , $A = diag (1, 2)$ , $B = diag (- 1 + ε, 0)$ , and $C = E_{11}$ . The Sylvester equation $A X + X B = C$ has a unique solution, its separation is $ε$ , and $X = ε^{- 1} E_{11}$ . Thus a residual of size $δ$ in the $E_{11}$ direction produces a solution error of size $δ / ε$ .

Facts & Assumptions

Given: The matrices and $ε$ in the Example.

[F1]

Sylvester operator and Frobenius separation number: $S_{A, B} (X) = A X + X B$ and $sep (A, - B) = σ_{\min} (S_{A, B})$ in the Frobenius inner product.

[F2]

A Sylvester equation is uniquely solvable exactly when the spectra are disjoint: Unique solvability holds exactly when $σ (A) \cap (- σ (B)) = \emptyset$ .

[F3]

Sylvester separation converts a residual into a forward error bound: $∥ X - \hat{X} ∥_{F} \leq ∥ C - (A \hat{X} + \hat{X} B) ∥_{F} / sep (A, - B)$ when the solution is unique.

Verification

1.1givenF1F2algebra

The eigenvalue sums are $ε, 1, 1 + ε, 2$ , all nonzero, so [F2] gives a unique solution. In the Frobenius-orthonormal matrix-unit basis, $S_{A, B} (E_{i j}) = (a_{i i} + b_{j j}) E_{i j}$ ; its singular values are therefore the four absolute eigenvalue sums, and [F1] gives $sep (A, - B) = ε$ .

2.1givenstep 1.1algebra

The $(i, j)$ equation is $(a_{i i} + b_{j j}) x_{i j} = c_{i j}$ . Since $C = E_{11}$ , the only nonzero entry is $x_{11} = 1 / ε$ , hence $X = ε^{- 1} E_{11}$ and $∥ X ∥_{F} = 1 / ε = 10^{6}$ .

3.1F1F3step 2.1algebra∎

If the right-hand side is perturbed by $δ E_{11}$ , the solution changes by $δ / ε E_{11}$ , whose Frobenius norm is $∣ δ ∣ / ε$ . This attains the separation factor in [F3], verifying the stated sensitivity exactly.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

One low-rank ADI step gives an exact residual certificate

Example

Let $A = - I_{2}$ , $B = e_{1} = (1, 0)^{T}$ , and take the single shift $p_{1} = - 1$ . One low-rank ADI step returns $V_{1} = - e_{1} / 2$ , $X_{1} = e_{1} e_{1}^{T} / 2$ , and $R_{1} = 0$ . The returned factor residual certificate is exact.

Facts & Assumptions

Given: The matrices and shift in the Example.

[F1]

Low-rank ADI for a continuous Lyapunov equation: $V_{1} = (A + p_{1} I)^{- 1} B$ , $c_{1} = - 2 Re (p_{1})$ , $X_{1} = c_{1} V_{1} V_{1}^{*}$ and $R_{1} = B + c_{1} V_{1}$ .

[F2]

Low-rank ADI has an exact residual factor identity: $A X_{1} + X_{1} A^{*} + B B^{*} = R_{1} R_{1}^{*}$ .

Verification

1.1givenF1algebra

$A + p_{1} I = - 2 I_{2}$ is invertible and $c_{1} = 2$ , so $V_{1} = (- 2 I_{2})^{- 1} e_{1} = - e_{1} / 2$ .

2.1F1step 1.1algebra

The updates in [F1] give $X_{1} = 2 (e_{1} e_{1}^{T} / 4) = e_{1} e_{1}^{T} / 2$ and $R_{1} = e_{1} + 2 (- e_{1} / 2) = 0$ .

3.1F2step 2.1algebra∎

Consequently $R_{1} R_{1}^{*} = 0$ , and directly $A X_{1} + X_{1} A^{*} + B B^{*} = - X_{1} - X_{1} + e_{1} e_{1}^{T} = 0$ , agreeing with [F2].