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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–nullity theorem ： ウィキペディア英語版 Rank–nullity theorem In mathematics, the rank–nullity theorem of linear algebra, in its simplest form, states that the rank and the nullity of a matrix add up to the number of columns of the matrix. Specifically, if ''A'' is an ''m''-by-''n'' matrix (with ''m'' rows and ''n'' columns) over some field, then〔, page 199.〕 :$\backslash operatorname(A)\; +\; \backslash operatorname(A)\; =\; n\; .$ This applies to linear maps as well. Let ''V'' and ''W'' be vector spaces over some field and let be a linear map. Then the rank of ''T'' is the dimension of the image of ''T'' and the nullity of ''T'' is the dimension of the kernel of ''T'', so we have :$\backslash operatorname(\backslash operatorname\; (T))\; +\; \backslash operatorname\; (\backslash operatorname\; (T))\; =\; \backslash operatorname\; (V)\; ,$ or, equivalently, :$\backslash operatorname(T)\; +\; \backslash operatorname(T)\; =\; \backslash operatorname(V).$ One can refine this statement (via the splitting lemma or the below proof) to be a statement about an isomorphism of spaces, not just dimensions. More generally, one can consider the image, kernel, coimage, and cokernel, which are related by the fundamental theorem of linear algebra. == Proofs == We give two proofs. The first proof uses notations for linear transformations, but can be easily adapted to matrices by writing , where A is . The second proof looks at the homogeneous system associated with an matrix A of rank ''r'' and shows explicitly that there exist a set of linearly independent solutions that span the null space of A. These proofs are also available in the book by Banerjee and Roy (2014) First proof: Suppose $\backslash \_m\backslash \}$ forms a basis of ''ker T''. We can extend this to form a basis of ''V'': $\backslash \_m,\; \backslash mathbf\_1,\; \backslash ldots,\; \backslash mathbf\_n\backslash \}$. Since the dimension of ker ''T'' is ''m'' and the dimension of ''V'' is , it suffices to show that the dimension of is ''n''. Let us see that $\backslash \_n\; \backslash \}$ is a basis of . Let ''v'' be an arbitrary vector in ''V''. There exist unique scalars such that: : $\backslash mathbf=a\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; a\_m\; \backslash mathbf\_m\; +\; b\_1\; \backslash mathbf\_1\; +\backslash cdots\; +\; b\_n\; \backslash mathbf\_n$ : $\backslash Rightarrow\; T\backslash mathbf\; =\; a\_1\; T\backslash mathbf\_1\; +\; \backslash cdots\; +\; a\_m\; T\backslash mathbf\_m\; +\; b\_1\; T\backslash mathbf\_1\; +\backslash cdots\; +\; b\_n\; T\backslash mathbf\_n$ : $\backslash Rightarrow\; T\backslash mathbf\; =\; b\_1\; T\backslash mathbf\_1\; +\; \backslash cdots\; +\; b\_n\; T\backslash mathbf\_n\; \backslash ;\; \backslash ;\; \backslash because\; T\backslash mathbf\_i\; =\; 0$ Thus, $\backslash \_n\; \backslash \}$ spans . We only now need to show that this list is not redundant; that is, that $\backslash \_n\; \backslash \}$ are linearly independent. We can do this by showing that a linear combination of these vectors is zero if and only if the coefficient on each vector is zero. Let: : $c\_1\; T\backslash mathbf\_1\; +\; \backslash cdots\; +\; c\_n\; T\backslash mathbf\_n\; =\; 0\; \backslash Leftrightarrow\; T(c\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; c\_n\; \backslash mathbf\_n)=0$ : $\backslash therefore\; c\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; c\_n\; \backslash mathbf\_n\; \backslash in\; \backslash operatorname\; \backslash ;\; T$ Then, since u''i'' span ker ''T'', there exists a set of scalars ''di'' such that: : $c\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; c\_n\; \backslash mathbf\_n\; =\; d\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; d\_m\; \backslash mathbf\_m$ But, since $\backslash \_m,\; \backslash mathbf\_1,\; \backslash ldots,\; \backslash mathbf\_n\backslash \}$ form a basis of ''V'', all ''ci'', ''di'' must be zero. Therefore, $\backslash \_n\; \backslash \}$ is linearly independent and indeed a basis of . This proves that the dimension of is ''n'', as desired. In more abstract terms, the map ''splits''. Second proof: Let A be an matrix with ''r'' linearly independent columns (i.e. rank of A is ''r''). We will show that: (i) there exists a set of linearly independent solutions to the homogeneous system , and (ii) that every other solution is a linear combination of these solutions. In other words, we will produce an matrix X whose columns form a basis of the null space of A. Without loss of generality, assume that the first ''r'' columns of A are linearly independent. So, we can write , where A1 is with ''r'' linearly independent column vectors and A2 is , each of whose columns are linear combinations of the columns of A1. This means that for some '' matrix B (see rank factorization) and, hence, . Let $\backslash displaystyle\; \backslash mathbf\; =$ \begin -\mathbf \\ \mathbf_ \end , where $\backslash mathbf\_$ is the identity matrix. We note that X is an matrix that satisfies :$$ \mathbf\mathbf = ()\begin -\mathbf \\ \mathbf_ \end = -\mathbf_1\mathbf + \mathbf_1\mathbf = \mathbf\; . Therefore, each of the columns of X are particular solutions of . Furthermore, the columns of X are linearly independent because will imply : :$\backslash mathbf\backslash mathbf\; =\; \backslash mathbf\; \backslash Rightarrow\; \backslash begin$ -\mathbf \\ \mathbf_ \end\mathbf = \mathbf \Rightarrow \begin -\mathbf\mathbf \\ \mathbf \end = \begin \mathbf \\ \mathbf \end \Rightarrow \mathbf = \mathbf\; . Therefore, the column vectors of X constitute a set of ''n'' − ''r'' linearly independent solutions for Ax = 0. We next prove that ''any'' solution of must be a linear combination of the columns of X For this, let $\backslash displaystyle\; \backslash mathbf\; =\; \backslash begin$ \mathbf_1 \\ \mathbf_2 \end be any vector such that . Note that since the columns of A1 are linearly independent, implies . Therefore, :$$ \mathbf\mathbf = \mathbf \Rightarrow ()\begin \mathbf_1 \\ \mathbf_2 \end = \mathbf \Rightarrow \mathbf_1(\mathbf_1 + \mathbf\mathbf_2) = \mathbf \Rightarrow \mathbf_1 + \mathbf\mathbf_2 = \mathbf \Rightarrow \mathbf_1 = -\mathbf\mathbf_2 : $\backslash Rightarrow\; \backslash mathbf\; =\; \backslash begin$ \mathbf_1 \\ \mathbf_2 \end = \begin -\mathbf \\ \mathbf_ \end\mathbf_2 = \mathbf\mathbf_2. This proves that any vector u that is a solution of must be a linear combination of the special solutions given by the columns of X. And we have already seen that the columns of X are linearly independent. Hence, the columns of X constitute a basis for the null space of A. Therefore, the nullity of A is . Since ''r'' equals rank of A, it follows that . QED. 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–nullity theorem」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.