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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 Leibniz formula (determinant) ： ウィキペディア英語版 Leibniz formula for determinants In algebra, the Leibniz formula, named in honor of Gottfried Leibniz, expresses the determinant of a square matrix in terms of permutations of the matrix elements. If ''A'' is an ''n''×''n'' matrix, where ''a''''i'',''j'' is the entry in the ''i''th row and ''j''th column of ''A'', the formula is :$\backslash det(A)\; =\; \backslash sum\_\; \backslash sgn(\backslash sigma)\; \backslash prod\_^n\; a\_,$ where sgn is the sign function of permutations in the permutation group ''S''''n'', which returns +1 and −1 for even and odd permutations, respectively. Another common notation used for the formula is in terms of the Levi-Civita symbol and makes use of the Einstein summation notation, where it becomes :$\backslash det(A)=\backslash epsilon^\_\backslash cdots\; \_,$ which may be more familiar to physicists. Directly evaluating the Leibniz formula from the definition requires $\backslash Omega(n!\; \backslash cdot\; n)$ operations in general—that is, a number of operations asymptotically proportional to ''n'' factorial—because ''n''! is the number of order-''n'' permutations. This is impractically difficult for large ''n''. Instead, the determinant can be evaluated in O(''n''3) operations by forming the LU decomposition $A\; =\; LU$ (typically via Gaussian elimination or similar methods), in which case $\backslash det\; A\; =\; (\backslash det\; L)\; (\backslash det\; U)$ and the determinants of the triangular matrices ''L'' and ''U'' are simply the products of their diagonal entries. (In practical applications of numerical linear algebra, however, explicit computation of the determinant is rarely required.) See, for example, . == Formal statement and proof == Theorem. There exists exactly one function :$F\; :\; M\_n\; (\backslash mathbb\; K)\; \backslash rightarrow\; \backslash mathbb\; K$ which is alternate multilinear w.r.t. columns and such that $F(I)\; =\; 1$. Proof. Uniqueness: Let $F$ be such a function, and let $A\; =\; (a\_i^j)\_^$ be an $n\; \backslash times\; n$ matrix. Call $A^j$ the $j$-th column of $A$, i.e. $A^j\; =\; (a\_i^j)\_$, so that $A\; =\; \backslash left(A^1,\; \backslash dots,\; A^n\backslash right).$ Also, let $E^k$ denote the $k$-th column vector of the identity matrix. Now one writes each of the $A^j$'s in terms of the $E^k$, i.e. :$A^j\; =\; \backslash sum\_^n\; a\_k^j\; E^k$. As $F$ is multilinear, one has :$$ \begin F(A)& = F\left(\sum_^n a_^1 E^, \dots, \sum_^n a_^n E^\right)\\ & = \sum_^n \left(\prod_^n a_^i\right) F\left(E^, \dots, E^\right). \end From alternation it follows that any term with repeated indices is zero. The sum can therefore be restricted to tuples with non-repeating indices, i.e. permutations: :$F(A)\; =\; \backslash sum\_\; \backslash left(\backslash prod\_^n\; a\_^i\backslash right)\; F(E^,\; \backslash dots\; ,\; E^).$ Because F is alternating, the columns $E$ can be swapped until it becomes the identity. The sign function $\backslash sgn(\backslash sigma)$ is defined to count the number of swaps necessary and account for the resulting sign change. One finally gets: :$$ \begin F(A)& = \sum_ \sgn(\sigma) \left(\prod_^n a_^i\right) F(I)\\ & = \sum_ \sgn(\sigma) \prod_^n a_^i \end as $F(I)$ is required to be equal to $1$. Therefore no function besides the function defined by the Leibniz Formula is a multilinear alternating function with $F\backslash left(I\backslash right)=1$. Existence: We now show that F, where F is the function defined by the Leibniz formula, has these three properties. Multilinear: :$$ \begin F(A^1, \dots, cA^j, \dots) & = \sum_ \sgn(\sigma) ca_^j\prod_^n a_^i\\ & = c \sum_ \sgn(\sigma) a_^j\prod_^n a_^i\\ &=c F(A^1, \dots, A^j, \dots)\\ \\ F(A^1, \dots, b+A^j, \dots) & = \sum_ \sgn(\sigma)\left(b_ + a_^j\right)\prod_^n a_^i\\ & = \sum_ \sgn(\sigma) \left( \left(b_\prod_^n a_^i\right) + \left(a_^j\prod_^n a_^i\right)\right)\\ & = \left(\sum_ \sgn(\sigma) b_\prod_^n a_^i\right) + \left(\sum_ \sgn(\sigma) \prod_^n a_^i\right)\\ &= F(A^1, \dots, b, \dots) + F(A^1, \dots, A^j, \dots)\\ \\ \end Alternating: :$$ \begin F(\dots, A^, \dots, A^, \dots) & = \sum_ \sgn(\sigma) \left(\prod_^n a_^i\right) a_^ a_^\\ \end For any $\backslash sigma\; \backslash in\; S\_n$ let $\backslash sigma\text{'}$ be the tuple equal to $\backslash sigma$ with the $j\_1$ and $j\_2$ indices switched. :$$ \begin F(A) & = \sum_)<\sigma(j_)}\left(\\ \end Thus if $A^\; =\; A^$ then $F(\backslash dots,\; A^,\; \backslash dots,\; A^,\; \backslash dots)=0$. Finally, $F(I)=1$: :$$ \begin\\ F(I) & = \sum_ \sgn(\sigma) \prod_^n I_^i\\ & = \sum_ \prod_^n I_^i\\ & = 1 \end Thus the only functions which are multilinear alternating with $F(I)=1$ are restricted to the function defined by the Leibniz formula, and it in fact also has these three properties. Hence the determinant can be defined as the only function :$\backslash det\; :\; M\_n\; (\backslash mathbb\; K)\; \backslash rightarrow\; \backslash mathbb\; K$ with these three properties. 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「Leibniz formula for determinants」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.