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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 Spectral radius ： ウィキペディア英語版 Spectral radius In mathematics, the spectral radius of a square matrix or a bounded linear operator is the supremum among the absolute values of the elements in its spectrum, which is sometimes denoted by ρ(·). ==Matrices== Let be the (real or complex) eigenvalues of a matrix . Then its spectral radius is defined as: :$\backslash rho(A)\; =\; \backslash max\; \backslash left\; \backslash .$ The following lemma shows a simple yet useful upper bound for the spectral radius of a matrix: :Lemma. Let with spectral radius and a consistent matrix norm ; then, for each : ::$\backslash rho(A)\backslash leq\; \backslash |A^k\backslash |^\}.$ ''Proof'': Let be an eigenvector-eigenvalue pair for a matrix ''A''. By the sub-multiplicative property of the matrix norm, we get: :$|\backslash lambda|^k\backslash |\backslash mathbf\backslash |\; =\; \backslash |\backslash lambda^k\; \backslash mathbf\backslash |\; =\; \backslash |A^k\; \backslash mathbf\backslash |\; \backslash leq\; \backslash |A^k\backslash |\backslash cdot\backslash |\backslash mathbf\backslash |$ and since we have :$|\backslash lambda|^k\; \backslash leq\; \backslash |A^k\backslash |$ and therefore :$\backslash rho(A)\backslash leq\; \backslash |A^k\backslash |^\}.$ The spectral radius is closely related to the behaviour of the convergence of the power sequence of a matrix; namely, the following theorem holds: :Theorem. Let with spectral radius ; then if and only if ::$\backslash lim\_\; A^k\; =\; 0.$ :Moreover, if , is not bounded for increasing values of . ''Proof.'' Assume the limit in question is zero, we will show that . Let be an eigenvector-eigenvalue pair for ''A''. Since we have: :$\backslash begin$ 0 &= \left(\lim_ A^k \right) \mathbf \\ &= \lim_ \left(A^k\mathbf \right ) \\ &= \lim_ \lambda^k\mathbf \\ &= \mathbf \lim_ \lambda^k \end and, since by hypothesis , we must have :$\backslash lim\_\backslash lambda^k\; =\; 0$ which implies |λ| < 1. Since this must be true for any eigenvalue λ, we can conclude ρ(''A'') < 1. Now assume the radius of is less than . From the Jordan normal form theorem, we know that for all , there exist with non-singular and block diagonal such that: :$A\; =\; VJV^$ with :$J=\backslash begin$ J_(\lambda_1) & 0 & 0 & \cdots & 0 \\ 0 & J_(\lambda_2) & 0 & \cdots & 0 \\ \vdots & \cdots & \ddots & \cdots & \vdots \\ 0 & \cdots & 0 & J_) & 0 \\ 0 & \cdots & \cdots & 0 & J_(\lambda_s) \end where :$J\_(\backslash lambda\_i)=\backslash begin$ \lambda_i & 1 & 0 & \cdots & 0 \\ 0 & \lambda_i & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i & 1 \\ 0 & 0 & \cdots & 0 & \lambda_i \end\in \mathbf^, 1\leq i\leq s. It is easy to see that :$A^k=VJ^kV^$ and, since is block-diagonal, :$J^k=\backslash begin$ J_^k(\lambda_1) & 0 & 0 & \cdots & 0 \\ 0 & J_^k(\lambda_2) & 0 & \cdots & 0 \\ \vdots & \cdots & \ddots & \cdots & \vdots \\ 0 & \cdots & 0 & J_) & 0 \\ 0 & \cdots & \cdots & 0 & J_^k(\lambda_s) \end Now, a standard result on the -power of an $m\_i\; \backslash times\; m\_i$ Jordan block states that, for $k\; \backslash geq\; m\_i-1$: :$J\_^k(\backslash lambda\_i)=\backslash begin$ \lambda_i^k & \lambda_i^ & \lambda_i^ & \cdots & \lambda_i^ \\ 0 & \lambda_i^k & \lambda_i^ & \cdots & \lambda_i^ \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_i^k & \lambda_i^ \\ 0 & 0 & \cdots & 0 & \lambda_i^k \end Thus, if then for all . Hence for all we have: :$\backslash lim\_J\_^k=0$ which implies :$\backslash lim\_\; J^k\; =\; 0.$ Therefore, :$\backslash lim\_A^k=\backslash lim\_VJ^kV^=V\; \backslash left\; (\backslash lim\_J^k\; \backslash right\; )V^=0$ On the other side, if $\backslash rho(A)>1$, there is at least one element in which doesn't remain bounded as k increases, so proving the second part of the statement. 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「Spectral radius」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.