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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 rank factorization ： ウィキペディア英語版 rank factorization Given an $m\; \backslash times\; n$ matrix $A$ of rank $r$, a rank decomposition or rank factorization of $A$ is a product $A=CF$, where $C$ is an $m\; \backslash times\; r$ matrix and $F$ is an $r\; \backslash times\; n$ matrix. Every finite-dimensional matrix has a rank decomposition: Let $A$ be an $m\backslash times\; n$ matrix whose column rank is $r$. Therefore, there are $r$ linearly independent columns in $A$; equivalently, the dimension of the column space of $A$ is $r$. Let $c\_1,c\_2,\backslash ldots,c\_r$ be any basis for the column space of $A$ and place them as column vectors to form the $m\backslash times\; r$ matrix $C\; =\; ()$. Therefore, every column vector of $A$ is a linear combination of the columns of $C$. To be precise, if $A\; =\; ()$ is an $m\backslash times\; n$ matrix with $a\_j$ as the $j$-th column, then :$a\_j\; =\; f\_c\_1\; +\; f\_c\_2\; +\; \backslash cdots\; +\; f\_c\_r,$ where $f\_$'s are the scalar coefficients of $a\_j$ in terms of the basis $c\_1,c\_2,\backslash ldots,c\_r$. This implies that $A\; =\; CF$, where $f\_$ is the $(i,j)$-th element of $F$. == rank(A) = rank(AT) == An immediate consequence of rank factorization is that the rank of $A$ is equal to the rank of its transpose $A^\backslash text$. Since the columns of $A$ are the rows of $A^\backslash text$, the column rank of $A$ equals its row rank. Proof: To see why this is true, let us first define rank to mean column rank. Since $A\; =\; CF$, it follows that $A^\backslash text\; =\; F^\backslash textC^\backslash text$. From the definition of matrix multiplication, this means that each column of $A^\backslash text$ is a linear combination of the columns of $F^\backslash text$. Therefore, the column space of $A^\backslash text$ is contained within the column space of $F^\backslash text$ and, hence, rank($A^\backslash text$) ≤ rank($F^\backslash text$). Now, $F^\backslash text$ is $n$×$r$, so there are $r$ columns in $F^\backslash text$ and, hence, rank($A^\backslash text$) ≤ $r$ = rank($A$). This proves that rank($A^\backslash text)$ ≤ rank($A$). Now apply the result to $A^\backslash text$ to obtain the reverse inequality: since $(A^\backslash text)^\backslash text$ = $A$, we can write rank($A$) = rank($(A^\backslash text)^\backslash text)$ ≤ rank($A^\backslash text$). This proves rank($A)$ ≤ rank($A^\backslash text$). We have, therefore, proved rank($A^\backslash text)$ ≤ rank($A$) and rank($A$) ≤ rank($A^\backslash text$), so rank($A$) = rank($A^\backslash text$). (Also see the first proof of column rank = row rank under rank). == Rank factorization from row echelon forms == In practice, we can construct one specific rank factorization as follows: we can compute $B$, the reduced row echelon form of $A$. Then $C$ is obtained by removing from $A$ all non-pivot columns, and $F$ by eliminating all zero rows of $B$. 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「rank factorization」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.