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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Sylvester equation ： ウィキペディア英語版 Sylvester equation In mathematics, in the field of control theory, a Sylvester equation is a matrix equation of the form:〔This equation is also commonly written in the equivalent form of ''AX-XB=C''.〕 :$A\; X\; +\; X\; B\; =\; C.$ Then given matrices ''A'',''B'', and ''C'', the problem is to find the possible matrices ''X'' that obey this equation. All matrices are assumed to have coefficients in the complex numbers. For the equation to make sense, the matrices must have appropriate sizes, for example they could all be square matrices of the same size. But more generally, we could take ''A'' and ''B'' must be square matrices of sizes ''n'' and ''m'' respectively, and then ''X'' and ''C'' both have ''n'' rows and ''m'' columns. A Sylvester equation has a unique solution for ''X'' exactly when there are no common eigenvalues of ''A'' and ''-B''. More generally, the equation ''AX''+''XB''=''C'' has been considered as an equation of bounded operators on a (possibly infinite-dimensional) Banach space. In this case, the condition for the uniqueness of a solution ''X'' is almost the same: There exists a unique solution ''X'' exactly when the spectra of ''A'' and ''-B'' are disjoint.〔Bhatia and Rosenthal, 1997〕 ==Existence and uniqueness of the solutions== Using the Kronecker product notation and the vectorization operator $\backslash operatorname$, we can rewrite Sylvester's equation in the form :$(I\_n\; \backslash otimes\; A\; +\; B^T\; \backslash otimes\; I\_n)\; \backslash operatornameX\; =\; \backslash operatornameC,$ where $I\_n$ is the $n\; \backslash times\; n$ identity matrix. In this form, the equation can be seen as a linear system of dimension $n^2\; \backslash times\; n^2$.〔 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.〕 Proposition. ''Given complex $n\backslash times\; n$ matrices $A$ and $B$, Sylvester's equation has a unique solution $X$ for all $C$ if and only if $A$ and $-B$ have no common eigenvalues.'' Proof. Consider the linear transformation $S:M\_n\backslash rightarrow\; M\_n$ given by $X\backslash mapsto\; AX+XB$. (i) Suppose that $A$ and $-B$ have no common eigenvalues. Then their characteristic polynomials $f(z)$ and $g(z)$ have highest common factor $1$. Hence there exist complex polynomials $p(z)$ and $q(z)$ such that $p(z)f(z)+q(z)g(z)=1$. By the Cayley–Hamilton theorem, $f(A)=0=g(-B)$; hence $g(A)q(A)=I$. Let $X$ be any solution of $S(X)=0$; so $AX=-XB$ and repeating this one sees that $X=q(A)g(A)X=q(A)Xg(-B)=0$. Hence by the rank plus nullity theorem $S$ is invertible, so for all $C$ there exists a unique solution $X$. (ii) Conversely, suppose that $s$ is a common eigenvalue of $A$ and $-B$. Note that $s$ is also an eigenvalue of the transpose $A^T$. Then there exist non-zero vectors $v$ and $w$ such that $A^Tw=sw$ and $Bv=-sv$. Choose $C$ such that $Cv=\backslash overline$, the vector whose entries are the complex conjugates of $w$. Then $AX+XB=C$ has no solution $X$, as is clear from the complex bilinear pairing ; the right-hand side is positive whereas the left is zero. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Sylvester equation」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.