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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 Schur complement ： ウィキペディア英語版 Schur complement In linear algebra and the theory of matrices, the Schur complement of a matrix block (i.e., a submatrix within a larger matrix) is defined as follows. Suppose ''A'', ''B'', ''C'', ''D'' are respectively ''p''×''p'', ''p''×''q'', ''q''×''p'' and ''q''×''q'' matrices, and ''D'' is invertible. Let : so that ''M'' is a (''p''+''q'')×(''p''+''q'') matrix. Then the Schur complement of the block ''D'' of the matrix ''M'' is the ''p''×''p'' matrix :$M/D\; =\; A-BD^C\; \backslash ,$ and the Schur complement of the block ''A'' of the matrix ''M'' is the ''q''×''q'' matrix :$M/A\; =\; D-CA^B.\; \backslash ,$ In the case that ''A'' or ''D'' is singular, the inverses on ''M/A'' and ''M/D'' can be replaced by a generalized inverse, yielding what is called the generalized Schur complement. The Schur complement is named after Issai Schur who used it to prove Schur's lemma, although it had been used previously. Emilie Haynsworth was the first to call it the ''Schur complement''.〔Haynsworth, E. V., "On the Schur Complement", ''Basel Mathematical Notes'', #BNB 20, 17 pages, June 1968.〕 The Schur complement is a key tool in the fields of numerical analysis, statistics and matrix analysis. ==Background== The Schur complement arises as the result of performing a block Gaussian elimination by multiplying the matrix ''M'' from the right with the "block lower triangular" matrix : Here ''Ip'' denotes a ''p''×''p'' identity matrix. After multiplication with the matrix ''L'' the Schur complement appears in the upper ''p''×''p'' block. The product matrix is :$$ \begin ML &= \left(A & B \\ C & D \end\right )\left(I_p & 0 \\ -D^C & I_q \end\right ) = \left(A-BD^C & B \\ 0 & D \end\right ) \\ &= \left(I_p & BD^ \\ 0 & I_q \end\right ) \left(A-BD^C & 0 \\ 0 & D \end\right ). \end This is analogous to an LDU decomposition. That is, we have shown that :$$ \begin \left(A & B \\ C & D \end\right ) &= \left(I_p & BD^ \\ 0 & I_q \end\right ) \left(A-BD^C & 0 \\ 0 & D \end\right ) \left(\begin I_p & 0 \\ D^C & I_q \end\right ), \end and inverse of ''M'' thus may be expressed involving ''D''−1 and the inverse of Schur's complement (if it exists) only as :$$ \begin & \right )^ = \left(\begin I_p & 0 \\ -D^C & I_q \end\right ) \left(\begin (A-BD^C)^ & 0 \\ 0 & D^ \end\right ) \left(\begin I_p & -BD^ \\ 0 & I_q \end\right ) \\() & = \left(\begin \left(A-B D^ C \right)^ & -\left(A-B D^ C \right)^ B D^ \\ -D^C\left(A-B D^ C \right)^ & D^+ D^ C \left(A-B D^ C \right)^ B D^ \end \right ). \end C.f. matrix inversion lemma which illustrates relationships between the above and the equivalent derivation with the roles of ''A'' and ''D'' interchanged. If ''M'' is a positive-definite symmetric matrix, then so is the Schur complement of ''D'' in ''M''. If ''p'' and ''q'' are both 1 (i.e. ''A'', ''B'', ''C'' and ''D'' are all scalars), we get the familiar formula for the inverse of a 2-by-2 matrix: : provided that ''AD'' − ''BC'' is non-zero. Moreover, the determinant of ''M'' is also clearly seen to be given by :$\backslash det(M)\; =\; \backslash det(D)\; \backslash det(A\; -\; BD^\; C)$ which generalizes the determinant formula for 2x2 matrices. 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「Schur complement」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.