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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 Jacobi's formula ： ウィキペディア英語版 Jacobi's formula In matrix calculus, Jacobi's formula expresses the derivative of the determinant of a matrix ''A'' in terms of the adjugate of ''A'' and the derivative of ''A''.〔, Part Three, Section 8.3〕 If ''A'' is a differentiable map from the real numbers to ''n'' × ''n'' matrices, :$\backslash frac\; \backslash det\; A(t)\; =\; \backslash mathrm\; \backslash left\; (\backslash mathrm(A(t))\; \backslash ,\; \backslash frac\backslash right\; )~.$ Equivalently, if ''dA'' stands for the differential of ''A'', the formula is :$d\; \backslash det\; (A)\; =\; \backslash mathrm\; (\backslash mathrm(A)\; \backslash ,\; dA).$ It is named after the mathematician C.G.J. Jacobi. ==Derivation== We first prove a preliminary lemma: Lemma. Let ''A'' and ''B'' be a pair of square matrices of the same dimension ''n''. Then :$\backslash sum\_i\; \backslash sum\_j\; A\_\; B\_\; =\; \backslash mathrm\; (A^\; B).$ ''Proof.'' The product ''AB'' of the pair of matrices has components :$(AB)\_\; =\; \backslash sum\_i\; A\_\; B\_.\backslash ,$ Replacing the matrix ''A'' by its transpose ''A''T is equivalent to permuting the indices of its components: :$(A^\; B)\_\; =\; \backslash sum\_i\; A\_\; B\_.$ The result follows by taking the trace of both sides: :$\backslash mathrm\; (A^\; B)\; =\; \backslash sum\_j\; (A^\; B)\_\; =\; \backslash sum\_j\; \backslash sum\_i\; A\_\; B\_\; =\; \backslash sum\_i\; \backslash sum\_j\; A\_\; B\_.\backslash \; \backslash square$ Theorem. (Jacobi's formula) For any differentiable map ''A'' from the real numbers to ''n'' × ''n'' matrices, : $d\; \backslash det\; (A)\; =\; \backslash mathrm\; (\backslash mathrm(A)\; \backslash ,\; dA).$ ''Proof.'' Laplace's formula for the determinant of a matrix ''A'' can be stated as :$\backslash det(A)\; =\; \backslash sum\_j\; A\_\; \backslash mathrm^\; (A)\_.$ Notice that the summation is performed over some arbitrary row ''i'' of the matrix. The determinant of ''A'' can be considered to be a function of the elements of ''A'': :$\backslash det(A)\; =\; F\backslash ,(A\_,\; A\_,\; \backslash ldots\; ,\; A\_,\; A\_,\; \backslash ldots\; ,\; A\_)$ so that, by the chain rule, its differential is :$d\; \backslash det(A)\; =\; \backslash sum\_i\; \backslash sum\_j\; .$ This summation is performed over all ''n''×''n'' elements of the matrix. To find ∂''F''/∂''A''''ij'' consider that on the right hand side of Laplace's formula, the index ''i'' can be chosen at will. (In order to optimize calculations: Any other choice would eventually yield the same result, but it could be much harder). In particular, it can be chosen to match the first index of ∂ / ∂''A''''ij'': :$\backslash mathrm^(A)\_\; \backslash over\; \backslash partial\; A\_\}\; =\; \backslash sum\_k\; ^(A)\_)\; \backslash over\; \backslash partial\; A\_\}$ Thus, by the product rule, :$\backslash over\; \backslash partial\; A\_\}\; \backslash mathrm^(A)\_\; +\; \backslash sum\_k\; A\_\; (A)\_\; \backslash over\; \backslash partial\; A\_\}.$ Now, if an element of a matrix ''A''''ij'' and a cofactor adjT(''A'')''ik'' of element ''A''''ik'' lie on the same row (or column), then the cofactor will not be a function of ''Aij'', because the cofactor of ''A''''ik'' is expressed in terms of elements not in its own row (nor column). Thus, :$(A)\_\; \backslash over\; \backslash partial\; A\_\}\; =\; 0,$ so :$^(A)\_\; \}.$ All the elements of ''A'' are independent of each other, i.e. :$\}\; =\; \backslash delta\_,$ where ''δ'' is the Kronecker delta, so :$^(A)\_\; \backslash delta\_\; =\; \backslash mathrm^(A)\_.$ Therefore, :$d(\backslash det(A))\; =\; \backslash sum\_i\; \backslash sum\_j\; \backslash mathrm^(A)\_\; \backslash ,d\; A\_,$ and applying the Lemma yields :$d(\backslash det(A))\; =\; \backslash mathrm(\backslash mathrm(A)\; \backslash ,dA).\backslash \; \backslash square$ 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「Jacobi's formula」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.