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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 Negative-definite matrix ： ウィキペディア英語版 Positive-definite matrix In linear algebra, a symmetric real matrix $M$ is said to be positive definite if the scalar $z^Mz$ is positive for every non-zero column vector $z$ of $n$ real numbers. Here $z^$ denotes the transpose of $z$. More generally, an Hermitian matrix $M$ is said to be positive definite if the scalar $z^Mz$ is real and positive for all non-zero column vectors $z$ of $n$ complex numbers. Here $z^$ denotes the conjugate transpose of $z$. The negative definite, positive semi-definite, and negative semi-definite matrices are defined in the same way, except that the expression $z^Mz$ or $z^Mz$ is required to be always negative, non-negative, and non-positive, respectively. Positive definite matrices are closely related to positive-definite symmetric bilinear forms (or sesquilinear forms in the complex case), and to inner products of vector spaces.〔(Stewart, J. (1976). Positive definite functions and generalizations, an historical survey. Rocky Mountain J. Math, 6(3). )〕 Some authors use more general definitions of "positive definite" that include some non-symmetric real matrices, or non-Hermitian complex ones. == Examples == *The identity matrix is positive definite. Seen as a real matrix, it is symmetric, and, for any non-zero column vector ''z'' with real entries ''a'' and ''b'', one has . Seen as a complex matrix, for any non-zero column vector ''z'' with complex entries ''a'' and ''b'' one has $z^$ *I z = \begin a^ * & b^ *\end \begin 1 & 0 \\ 0 & 1\end \begin a \\ b\end=a^ *a +b^ *b = |a|^2 + |b|^2. Either way, the result is positive since ''z'' is not the zero vector (that is, at least one of ''a'' and ''b'' is not zero). * The real symmetric matrix :: :is positive definite since for any non-zero column vector ''z'' with entries ''a'', ''b'' and ''c'', we have :: &= 2^2 - 2ab + 2^2 - 2bc + 2^2 \\ &= ^2+(a - b)^ + (b - c)^+^2 \end :This result is a sum of squares, and therefore non-negative; and is zero only if ''a'' = ''b'' = ''c'' = 0, that is, when ''z'' is zero. *The real symmetric matrix :: :is not positive definite. If ''z'' is the vector $\backslash begin\; -1\backslash \backslash \; 1\backslash end$, one has *For any real non-singular matrix $A$, the product $A^T\; A$ is a positive definite matrix. A simple proof is that for any non-zero vector $z$, the condition $z^T\; A^T\; A\; z\; =\; \backslash |\; A\; z\; \backslash |\_2^2\; >\; 0,$ since the non-singularity of matrix $A$ means that $Az\; \backslash neq\; 0.$ The examples ''M'' and ''N'' above show that a matrix in which some elements are negative may still be positive-definite, and conversely a matrix whose entries are all positive may not be positive definite. 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「Positive-definite matrix」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.