设A为n阶矩阵,且每一行元素之和为a,证明A^m的每一行元素之和为a^m
每一行元素之和为a则A(1,1...1)T=a(1,1...1)T
所以A^m(1,1...1)T=a^m(1,1...1)T即
A^m的每一行元素之和为a^m
(1,1...1)T是个列向量，每个元素都是1
A乘以这个列向量得出的就是A的每行元素和
绛旓細鎵浠^m(1,1...1)T=a^m(1,1...1)T鍗 A^m鐨勬瘡涓琛屽厓绱犱箣鍜屼负a^m (1,1...1)T鏄釜鍒楀悜閲忥紝姣忎釜鍏冪礌閮芥槸1 A涔樹互杩欎釜鍒楀悜閲忓緱鍑虹殑灏辨槸A鐨姣忚鍏冪礌鍜
绛旓細鎵浠 a^m 鏄 (A^T)^m 鐨勭壒寰佸, (1,1,...,1) 鏄(A^T)^m鐨勫睘浜庣壒寰佸糰^m鐨勭壒寰佸悜閲 鍥犱负 (A^T)^m = (A^m)^T 鎵浠ユ湁 (A^m)^T (1,1,...,1)^T = a^m (1,1,...,1)^T 鍗虫湁 A^m 鐨勬瘡鍒鍏冪礌涔鍜屼负甯告暟 a^m....
绛旓細鍥犱负姣忚鍏冪礌涔鍜屼负0 鎵浠=t(1,1,...,1)^T鏄竴鏃忚В 鍙堝洜涓簉(A)=N-1 鎵浠x=0鐨勮В鏄竴缁村瓙绌洪棿锛坸=t(1,1,...,1)^T灏辨槸涓缁寸殑锛夋墍浠ラ氳В涓簒=t(1,1,...,1)^T
绛旓細n闃剁煩闃礎鐨鍚勮鍏冪礌涔鍜岄兘涓3 閭ｄ箞鏄剧劧A涔樹互(1锛1锛1鈥1)^T 鍗冲緱鍒扮殑鐗瑰緛鍚戦噺姣忎釜鍏冪礌 閮芥槸鍚勮鍏冪礌鐩稿姞锛屼负3 鎵浠(1锛1锛1鈥1)^T=3(1锛1锛1鈥1)^T 浜庢槸A鐨勪竴涓壒寰佸间负3 鐩稿簲鐨勭壒寰佸悜閲忓氨鏄(1锛1锛1鈥1)^T
绛旓細璁剧煩闃涓篈锛岄渶瑕佽瘉鏄庡瓨鍦ㄩ潪闆跺悜閲弜锛屼娇寰桝x = ax锛屽洜涓篈琛屽拰鐩稿悓锛屼笖琛屽拰涓篴锛屽彇x = [1 1 ... 1]' 鍏冪礌鍏ㄤ负1鐨勫垪鍚戦噺锛屽垯鏄剧劧Ax = ax锛屾墍浠鏄壒寰佸笺傞潪闆秐缁村垪鍚戦噺x绉颁负鐭╅樀A鐨勫睘浜庯紙瀵瑰簲浜庯級鐗瑰緛鍊糾鐨勭壒寰佸悜閲忔垨鏈緛鍚戦噺锛岀畝绉癆鐨勭壒寰佸悜閲忔垨A鐨勬湰寰佸悜閲忋璁続鏄痭闃舵柟闃碉紝濡傛灉瀛樺湪...
绛旓細瑙ｇ瓟杩囩▼濡備笅锛n闃剁煩闃A鐨勫悇琛鍏冪礌涔鍜屽潎涓洪浂锛岃鏄庯紙1锛1锛屸︼紝1锛塗锛坣涓1鐨勫垪鍚戦噺锛変负Ax=0鐨勪竴涓В銆傜敱浜嶢鐨勭З涓猴細n-1锛屼粠鑰屽熀纭瑙ｇ郴鐨勭淮搴︿负锛歯-r锛圓锛夛紝鏁匒鐨勫熀纭瑙ｇ郴鐨勭淮搴︿负1銆傜敱浜庯紙1锛1锛屸︼紝1锛塗鏄柟绋嬬殑涓涓В锛屼笉涓0锛屾墍浠x=0鐨勯氳В涓猴細k锛1锛1锛屸︼紝1锛塗銆
绛旓細璇佹槑: 璁 x=(1,1,...,1)^T.鐢卞凡鐭鐨姣忎竴琛屽厓绱犱箣鍜屼负c 鎵浠 Ax = (c,c,...,c)^T = cx.鎵浠 A^-1Ax = cA^-1x 鍗 x = cA^-1x 鎵浠 A^-1x = (1/c)x.--娉: 鍥犱负A鍙, 鏁卌鈮0 鎵浠^-1鐨姣忎竴琛屽厓绱犱箣鍜屼负 1/c.
绛旓細A*涓嶅彲閫,鍗 AA*=|A|E=O 浠庤 r(A)+r(A*),9,pluto1981 涓炬姤 濡傛灉 r(A)涓炬姤 yinyan102102 A*鏄敱A鐨勪唬鏁颁綑瀛愬紡鏋勬垚鐨 姣忎釜閮芥槸n-1闃剁殑琛屽垪寮忥紝鑰 r(A) 鍟婏紝瑙夋偀浜!!璋㈣阿鍟!!!,渚濅簯鍌嶉 骞艰嫍 鍏卞洖绛斾簡27涓棶棰 鍚慣A鎻愰棶 涓炬姤 璁続鏄痭闃剁煩闃碉紝瀵逛簬榻愭绾挎ф柟绋嬬粍AX=0...
绛旓細鑰冭檻鍒楀悜閲弜=(1, 1, ..., 1)瀹冨拰璇鐭╅樀鐨勪箻绉槸(a,a,...,a)瀹冩弧瓒矨x = ax锛屽洜姝a鏄鐗瑰緛鍊硷紝x鏄壒寰佸悜閲
绛旓細n闃剁煩闃A鐨勫悇琛鍏冪礌涔鍜屽潎涓洪浂锛岃鏄(1,1,...,1)T (n涓1鐨勫垪鍚戦噺) 涓篈x=0鐨勪竴涓В锛岀敱浜嶢鐨勭З涓簄-1,鏁匒鐨勫熀纭瑙ｇ郴鐨勭淮搴︿负1锛屼簬鏄疉x=0鐨勯氳В涓簁(1,1,...,1)T
