韦斯特方程

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Fernando（讨论 | 贡献）2021年9月5日 (日) 12:38的版本 （审校第一部分）
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西尔韦斯特方程是控制论中的矩阵方程，形式如下[1]:

[math]\displaystyle{ A X + X B = C. }[/math]

其中A，B，C为已知的矩阵，问题是要找到能够满足方程的矩阵X。所有矩阵的系数都是复数。为了使方程有意义，矩阵的行和列需要满足一定条件，A 和 B 都要是方阵，大小分别是n和m，而X和C要是n行m列的矩阵，n和m也可以相等，四个矩阵都是大小相同的方阵。

当且仅当 A 和-b 没有共同的本征值时，西尔韦斯特有唯一解 X 。更一般地，方程 AX + XB = c 也可以视为(可能无限维中)巴拿赫空间中有界算子的方程。在这种情况下，此情形下，有唯一解 X 的充分必要条件几乎相同: A 和-B 的谱不相交[2]。

Existence and uniqueness of the solutions

Using the Kronecker product notation and the vectorization operator [math]\displaystyle{ \operatorname{vec} }[/math], we can rewrite Sylvester's equation in the form

[math]\displaystyle{ (I_m \otimes A + B^T \otimes I_n) \operatorname{vec}X = \operatorname{vec}C, }[/math]

where [math]\displaystyle{ A }[/math] is of dimension [math]\displaystyle{ n\! \times\! n }[/math], [math]\displaystyle{ B }[/math] is of dimension [math]\displaystyle{ m\!\times\!m }[/math], [math]\displaystyle{ X }[/math] of dimension [math]\displaystyle{ n\!\times\!m }[/math] and [math]\displaystyle{ I_k }[/math] is the [math]\displaystyle{ k \times k }[/math] identity matrix. In this form, the equation can be seen as a linear system of dimension [math]\displaystyle{ mn \times mn }[/math].[3]

Using the Kronecker product notation and the vectorization operator \operatorname{vec}, we can rewrite Sylvester's equation in the form

(I_m \otimes A + B^T \otimes I_n) \operatorname{vec}X = \operatorname{vec}C,

where A is of dimension n\! \times\! n, B is of dimension m\!\times\!m, X of dimension n\!\times\!m and I_k is the k \times k identity matrix. In this form, the equation can be seen as a linear system of dimension mn \times mn.However, rewriting the equation in this form is not advised for the numerical solution since this version is costly to solve and can be ill-conditioned.

使用克罗内克积符号和向量运算符操作符{ vec } ，我们可以以下形式重写 Sylvester 的方程: (i m 乘以 a + b ^ t 乘以 i n)操作符名{ vec } x = 操作符名{ vec } c，其中 a 是 n 维数！时代周刊！N，b 是维数 m! 乘以! m，x 是维数 n! 乘以! m，i _ k 是 k 乘以 k 的单位矩阵。在这种形式下，该方程可以看作是一个维数为 mn 乘以 mn 的线性方程组。然而，重写这种形式的方程不建议数值解，因为这个版本是昂贵的解决，可以病态。

Theorem. Given matrices [math]\displaystyle{ A\in \mathbb{C}^{n\times n} }[/math] and [math]\displaystyle{ B\in \mathbb{C}^{m\times m} }[/math], the Sylvester equation [math]\displaystyle{ AX+XB=C }[/math] has a unique solution [math]\displaystyle{ X\in \mathbb{C}^{n\times m} }[/math] for any [math]\displaystyle{ C\in\mathbb{C}^{n\times m} }[/math] if and only if [math]\displaystyle{ A }[/math] and [math]\displaystyle{ -B }[/math] do not share any eigenvalue.

Theorem. Given matrices A\in \mathbb{C}^{n\times n} and B\in \mathbb{C}^{m\times m}, the Sylvester equation AX+XB=C has a unique solution X\in \mathbb{C}^{n\times m} for any C\in\mathbb{C}^{n\times m} if and only if A and -B do not share any eigenvalue.

定理。给定矩阵 a 在数学上{ c } ^ { n 次 n }和 b 在数学上{ c } ^ { m 次 m } ，Sylvester 方程 ax + xb = c 对任意 c 在数学上{ c } ^ { n 次 m }有唯一解 x 当且仅当 a 和-b 不共享任何特征值。

Proof. The equation [math]\displaystyle{ AX+XB=C }[/math] is a linear system with [math]\displaystyle{ mn }[/math] unknowns and the same amount of equations. Hence it is uniquely solvable for any given [math]\displaystyle{ C }[/math] if and only if the homogeneous equation [math]\displaystyle{ AX+XB=0 }[/math] admits only the trivial solution [math]\displaystyle{ 0 }[/math].

Proof. The equation AX+XB=C is a linear system with mn unknowns and the same amount of equations. Hence it is uniquely solvable for any given C if and only if the homogeneous equation

AX+XB=0

admits only the trivial solution 0.

证据。方程 ax + xb = c 是一个线性方程组，其中有许多未知数和等量的方程。因此，对于任意给定的 c，它是唯一可解的当且仅当齐次方程 ax + xb = 0仅存在平凡解0。

(i) Assume that [math]\displaystyle{ A }[/math] and [math]\displaystyle{ -B }[/math] do not share any eigenvalue. Let [math]\displaystyle{ X }[/math] be a solution to the abovementioned homogeneous equation. Then [math]\displaystyle{ AX=X(-B) }[/math], which can be lifted to [math]\displaystyle{ A^kX = X(-B)^k }[/math] for each [math]\displaystyle{ k \ge 0 }[/math] by mathematical induction. Consequently, [math]\displaystyle{ p(A) X = X p(-B) }[/math] for any polynomial [math]\displaystyle{ p }[/math]. In particular, let [math]\displaystyle{ p }[/math] be the characteristic polynomial of [math]\displaystyle{ A }[/math]. Then [math]\displaystyle{ p(A)=0 }[/math] due to the Cayley-Hamilton theorem; meanwhile, the spectral mapping theorem tells us [math]\displaystyle{ \sigma(p(-B)) = p(\sigma(-B)), }[/math] where [math]\displaystyle{ \sigma(\cdot) }[/math] denotes the spectrum of a matrix. Since [math]\displaystyle{ A }[/math] and [math]\displaystyle{ -B }[/math] do not share any eigenvalue, [math]\displaystyle{ p(\sigma(-B)) }[/math] does not contain zero, and hence [math]\displaystyle{ p(-B) }[/math] is nonsingular. Thus [math]\displaystyle{ X= 0 }[/math] as desired. This proves the "if" part of the theorem.

(i) Assume that A and -B do not share any eigenvalue. Let X be a solution to the abovementioned homogeneous equation. Then AX=X(-B), which can be lifted to

A^kX = X(-B)^k

for each k \ge 0 by mathematical induction. Consequently,

p(A) X = X p(-B)

for any polynomial p. In particular, let p be the characteristic polynomial of A. Then p(A)=0 due to the Cayley-Hamilton theorem; meanwhile, the spectral mapping theorem tells us

\sigma(p(-B)) = p(\sigma(-B)),

where \sigma(\cdot) denotes the spectrum of a matrix. Since A and -B do not share any eigenvalue, p(\sigma(-B)) does not contain zero, and hence p(-B) is nonsingular. Thus X= 0 as desired. This proves the "if" part of the theorem.

(i)假设 a 和-b 不共享任何特征值。设 x 是上述齐次方程的解。然后 AX = x (- b) ，它可以被数学归纳法提升到 a ^ kX = x (- b) ^ k。因此，对于任意多项式 p，p (a) x = xp (- b) ，特别地，设 p 是 a 的特征多项式。然后由于 Cayley-Hamilton 定理 p (a) = 0，同时谱映射定理告诉我们 sigma (p (- b)) = p (sigma (- b)) ，其中 sigma (cdot)表示矩阵的谱。由于 a 和-b 不共享任何特征值，p (sigma (- b))不包含零，因此 p (- b)是非奇异的。因此 x = 0。这证明了定理的“如果”部分。

(ii) Now assume that [math]\displaystyle{ A }[/math] and [math]\displaystyle{ -B }[/math] share an eigenvalue [math]\displaystyle{ \lambda }[/math]. Let [math]\displaystyle{ u }[/math] be a corresponding right eigenvector for [math]\displaystyle{ A }[/math], [math]\displaystyle{ v }[/math] be a corresponding left eigenvector for [math]\displaystyle{ -B }[/math], and [math]\displaystyle{ X=u{v}^* }[/math]. Then [math]\displaystyle{ X\neq 0 }[/math], and [math]\displaystyle{ AX+XB = A(uv^*)-(uv^*)(-B) = \lambda uv^*-\lambda uv^* = 0. }[/math] Hence [math]\displaystyle{ X }[/math] is a nontrivial solution to the aforesaid homogeneous equation, justifying the "only if" part of the theorem. Q.E.D.

(ii) Now assume that A and -B share an eigenvalue \lambda. Let u be a corresponding right eigenvector for A, v be a corresponding left eigenvector for -B, and X=u{v}^*. Then X\neq 0, and

AX+XB = A(uv^*)-(uv^*)(-B) = \lambda uv^*-\lambda uv^* = 0.

Hence X is a nontrivial solution to the aforesaid homogeneous equation, justifying the "only if" part of the theorem. Q.E.D.

(ii)现在假设 a 和-b 共享一个特征值 λ。设 u 是 a 的对应右特征向量，v 是-b 的对应左特征向量，x = u { v } ^ * 。然后 x neq 0，ax + xb = a (uv ^ *)-(uv ^ *)(- b) = lambda uv ^ *-lambda uv ^ * = 0。因此，x 是上述齐次方程的非平凡解，证明了定理的“只有当”部分。Q.e.d.

As an alternative to the spectral mapping theorem, the nonsigularity of [math]\displaystyle{ p(-B) }[/math] in part (i) of the proof can also be demonstrated by the Bézout's identity for coprime polynomials. Let [math]\displaystyle{ q }[/math] be the characteristic polynomial of [math]\displaystyle{ -B }[/math]. Since [math]\displaystyle{ A }[/math] and [math]\displaystyle{ -B }[/math] do not share any eigenvalue, [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are coprime. Hence there exist polynomials [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math] such that [math]\displaystyle{ p(z)f(z)+q(z)g(z)\equiv 1 }[/math]. By the Cayley–Hamilton theorem, [math]\displaystyle{ q(-B)=0 }[/math]. Thus [math]\displaystyle{ p(-B)f(-B)=I }[/math], implying that [math]\displaystyle{ p(-B) }[/math] is nonsigular.

As an alternative to the spectral mapping theorem, the nonsigularity of p(-B) in part (i) of the proof can also be demonstrated by the Bézout's identity for coprime polynomials. Let q be the characteristic polynomial of -B. Since A and -B do not share any eigenvalue, p and q are coprime. Hence there exist polynomials f and g such that p(z)f(z)+q(z)g(z)\equiv 1. By the Cayley–Hamilton theorem, q(-B)=0. Thus p(-B)f(-B)=I, implying that p(-B) is nonsigular.

作为谱映射定理的另一种形式，证明的第(i)部分 p (- b)的非图形性也可以用贝祖特的互素多项式恒等式来证明。设 q 是-b 的特征多项式。由于 a 和-b 不共享任何特征值，所以 p 和 q 是互素。因此存在多项式 f 和 g，使得 p (z) f (z) + q (z) g (z)等于1。利用 Cayley-Hamilton 定理，q (- b) = 0。因此 p (- b) f (- b) = i，意味着 p (- b)是非奇异的。

The theorem remains true for real matrices with the caveat that one considers their complex eigenvalues. The proof for the "if" part is still applicable; for the "only if" part, note that both [math]\displaystyle{ \mathrm{Re}(uv^*) }[/math] and [math]\displaystyle{ \mathrm{Im}(uv^*) }[/math] satisfy the homogenous equation [math]\displaystyle{ AX+XB=0 }[/math], and they cannot be zero simultaneously.

The theorem remains true for real matrices with the caveat that one considers their complex eigenvalues. The proof for the "if" part is still applicable; for the "only if" part, note that both \mathrm{Re}(uv^*) and \mathrm{Im}(uv^*) satisfy the homogenous equation AX+XB=0, and they cannot be zero simultaneously.

这个定理对实矩阵仍然成立，但需要注意的是考虑它们的复特征值。“如果”部分的证明仍然适用; 对于“只有如果”部分，请注意 mathrum { Re }(uv ^ *)和 mathrum { Im }(uv ^ *)都满足齐次方程 ax + xb = 0，它们不能同时为零。

Roth's removal rule

Roth's removal rule

= Roth’s removal rule =

Given two square complex matrices A and B, of size n and m, and a matrix C of size n by m, then one can ask when the following two square matrices of size n + m are similar to each other: [math]\displaystyle{ \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} }[/math] and [math]\displaystyle{ \begin{bmatrix} A & 0 \\0&B \end{bmatrix} }[/math]. The answer is that these two matrices are similar exactly when there exists a matrix X such that AX − XB = C. In other words, X is a solution to a Sylvester equation. This is known as Roth's removal rule.[4]

Given two square complex matrices A and B, of size n and m, and a matrix C of size n by m, then one can ask when the following two square matrices of size n + m are similar to each other: \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} and \begin{bmatrix} A & 0 \\0&B \end{bmatrix}. The answer is that these two matrices are similar exactly when there exists a matrix X such that AX − XB = C. In other words, X is a solution to a Sylvester equation. This is known as Roth's removal rule.

给定两个 n 和 m 的方阵复矩阵 a 和 b，以及一个 n × m 的矩阵 c，我们可以问下列两个 n + m 的方阵什么时候相似: 开始{ bmatrix } a & c0 & b 结束{ bmatrix } ，开始{ bmatrix } a & 0 & b 结束{ bmatrix }。答案是当存在一个矩阵 x 使得 AX-XB = c 时，这两个矩阵是完全相似的。换句话说，x 是 Sylvester 方程的解。这就是众所周知的罗斯免职规则。

One easily checks one direction: If AX − XB = C then

[math]\displaystyle{ \begin{bmatrix}I_n & X \\ 0 & I_m \end{bmatrix} \begin{bmatrix} A&C\\0&B \end{bmatrix} \begin{bmatrix} I_n & -X \\ 0& I_m \end{bmatrix} = \begin{bmatrix} A&0\\0&B \end{bmatrix}. }[/math]

Roth's removal rule does not generalize to infinite-dimensional bounded operators on a Banach space.[5]

One easily checks one direction: If AX − XB = C then

\begin{bmatrix}I_n & X \\ 0 & I_m \end{bmatrix} \begin{bmatrix} A&C\\0&B \end{bmatrix} \begin{bmatrix} I_n & -X \\ 0& I_m \end{bmatrix} = \begin{bmatrix} A&0\\0&B \end{bmatrix}.

Roth's removal rule does not generalize to infinite-dimensional bounded operators on a Banach space.Bhatia and Rosenthal, p.3

一个很容易检查的方向: 如果 AX-XB = c，那么开始{ bmatrix } i _ n & x0 & i _ m end { bmatrix } begin { bmatrix } a & c0 & b end { bmatrix } begin { bmatrix } i _ n &-x0 & i _ m end { bmatrix } = begin { bmatrix } a & 0 & b end { bmatrix }。Roth 的移除规则不能推广到 Banach 空间上的无限维有界算子

Numerical solutions

A classical algorithm for the numerical solution of the Sylvester equation is the Bartels–Stewart algorithm, which consists of transforming [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] into Schur form by a QR algorithm, and then solving the resulting triangular system via back-substitution. This algorithm, whose computational cost is [math]\displaystyle{ \mathcal{O}(n^3) }[/math] arithmetical operations,[citation needed] is used, among others, by LAPACK and the lyap function in GNU Octave.[6] See also the sylvester function in that language.[7][8] In some specific image processing application, the derived Sylvester equation has a closed form solution.[9]

A classical algorithm for the numerical solution of the Sylvester equation is the Bartels–Stewart algorithm, which consists of transforming A and B into Schur form by a QR algorithm, and then solving the resulting triangular system via back-substitution. This algorithm, whose computational cost is \mathcal{O}(n^3) arithmetical operations, is used, among others, by LAPACK and the lyap function in GNU Octave. See also the sylvester function in that language.The syl command is deprecated since GNU Octave Version 4.0 In some specific image processing application, the derived Sylvester equation has a closed form solution.

Sylvester 方程数值解的一个经典算法是 Bartels-Stewart 算法，该算法通过 QR 算法将 a 和 b 转化为 Schur 形式，然后通过反代换求解得到三角形方程组。该算法的计算代价是数学{ o }(n ^ 3)算术运算，其中包括 LAPACK ack 和 GNU Octave 中的 lyap 函数。请参阅该语言中的 sylvester 函数。自 GNU Octave Version 4.0以来，不推荐使用 < code > syl 命令。在某些特定的图像处理应用程序中，导出的 Sylvester 方程有一个封闭的解。

See also

  • Lyapunov equation
  • Algebraic Riccati equation

方程

  • 代数 Riccati方程

Notes

  1. ↑ This equation is also commonly written in the equivalent form of AX − XB = C.
  2. ↑ Bhatia and Rosenthal, 1997
  3. ↑ However, rewriting the equation in this form is not advised for the numerical solution since this version is costly to solve and can be ill-conditioned.
  4. ↑ Gerrish, F; Ward, A.G.B (Nov 1998). "Sylvester's matrix equation and Roth's removal rule". The Mathematical Gazette. 82 (495): 423–430. doi:10.2307/3619888. JSTOR 3619888.
  5. ↑ Bhatia and Rosenthal, p.3
  6. ↑ "Function Reference: Lyap".
  7. ↑ "Functions of a Matrix (GNU Octave (version 4.4.1))".
  8. ↑ The syl command is deprecated since GNU Octave Version 4.0
  9. ↑ Wei, Q.; Dobigeon, N.; Tourneret, J.-Y. (2015). "Fast Fusion of Multi-Band Images Based on Solving a Sylvester Equation". IEEE. 24 (11): 4109–4121. arXiv:1502.03121. Bibcode:2015ITIP...24.4109W. doi:10.1109/TIP.2015.2458572. PMID 26208345.

References


External links

  • Online solver for arbitrary sized matrices.
  • Mathematica function to solve the Sylvester equation
  • MATLAB function to solve the Sylvester equation


  • 任意大小矩阵的在线求解器。
  • Mathematica 函数求解 Sylvester 方程
  • MATLAB 函数求解 Sylvester 方程

Category:Matrices Category:Control theory

范畴: 矩阵范畴: 控制论


This page was moved from wikipedia:en:Sylvester equation. Its edit history can be viewed at 韦斯特方程/edithistory

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