都是收敛的，于是矩阵级数 绝对收敛。 反之，若矩阵级数 41 绝对收敛，则对每一对i,方都有
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a,=la,+a,+…+a,+ k-0 <0 于是 211芝2叭-22， :1 11 即4收敛。根据矩阵范数等价性定理知结 论对任何一种范数都正确
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关于矩阵级数运算有下面的性质： 设至m九 宫-8 ,则 () aA+BB)=aA+BB,a,peC k=0 ②如果立40 收敛（绝对收敛），则 立Q=空0收致（绝对收敛）。 =0 3)若】 立g均绝对收文。则 k=0 4xB=AB(绝对收敛) 0 This docu at is produced by trial ver f Print2Flash.Visit www.print2flash.com for m
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定义设A=(a)∈Cm”,称形如 erAt=en1teAtted 的矩阵级数为矩阵幂级数。 定理3.6设幂级数 三c,的收敛半径为八A 为n阶方阵。若p(A)<r，则矩阵幂级数 之c4绝对收敛：若p(0>r,则三4 发散。 by tria
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证明设A的Jordan标准形为 J=diag(J1(）,J2(22),…,J,(,) 其中 21 J,(2)= (i=1,2,…,r） 于是 Mi d,xd A=Pdiag(J(),J(2),…,J(2)P1 This doc is pro duced by trial ver w.print2flash.com
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