由 亂解一通 於 星期二 四月 18, 2006 12:48 pm
#ed_op#DIV#ed_cl#(a) rank(A)= n-1 if and only if rank(adjA)=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#Pf (a)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1.先證明 A,B為n階方陣 iff rank(AB) ≥ rank(A)+rank(B) - n #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#Note:事實上若 A 是 s by n ,B是 n by t也會成立。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.由rank(A) = n-1 保證 det(A) = 0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故A(adjA) = det(A) I = 0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以rank[A(adjA)] = 0 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#利用上面定理可以知道 0 ≥ n-1 + rank(adjA) -n <=> rank(adjA) ≤1 ..... (1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3.又由於rank(A) =n-1 可以知道至少在Algebric Complementary Minors中有一個entry不為0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#<=> rank(adjA) ≥ 1.....(2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#4. By (1),(2) 可以知道成立的條件僅在 rank(adjA) =1。 證畢!#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#