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由 cloudsea 於 星期二 四月 18, 2006 2:14 am
謝謝樓上的解惑...^^
由 亂解一通 於 星期一 四月 17, 2006 3:26 pm
#ed_op#DIV#ed_cl#設B為n階對角線矩陣，不失一般性，設每個對角元素皆不相同。即[b#ed_op#SUB#ed_cl#ij#ed_op#/SUB#ed_cl#] = 0 ,i≠j ，餘不為0且不相等。 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#另設A為n階任意矩陣，以[a#ed_op#SUB#ed_cl#ij#ed_op#/SUB#ed_cl#]表示。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#並且已知AB=BA#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由矩陣乘法知道，要使上面條件滿足必須要 a#ed_op#SUB#ed_cl#ij#ed_op#/SUB#ed_cl#b#ed_op#SUB#ed_cl#jj#ed_op#/SUB#ed_cl# = b#ed_op#SUB#ed_cl#ii#ed_op#/SUB#ed_cl#a#ed_op#SUB#ed_cl#ij #ed_op#/SUB#ed_cl#for all i , j #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以，可得a#ed_op#SUB#ed_cl#ij#ed_op#/SUB#ed_cl#(b#ed_op#SUB#ed_cl#ii #ed_op#/SUB#ed_cl#- b#ed_op#SUB#ed_cl#jj#ed_op#/SUB#ed_cl#) = 0 ∀i , j #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#已知B上的對角元素不相等(當i , j 不相等)， 所以a#ed_op#SUB#ed_cl#ij #ed_op#/SUB#ed_cl#= 0 ∀i ≠j #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以A也為n階對角線矩陣。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#
[問題]線代問題...希望高手幫忙...謝謝
由 cloudsea 於 星期一 四月 17, 2006 1:13 pm
#ed_op#P#ed_cl#Two matrices A and B are said to commute if AB=BA .Find the set of all A 屬於M#ed_op#SUB#ed_cl#n*n#ed_op#/SUB#ed_cl# whcih commute with all the diagonal matrices in M#ed_op#SUB#ed_cl#n*n#ed_op#/SUB#ed_cl# 我想到的想法是AB=BA所以可以同步對角化...但是做到後來得到D1*D2=D2*D1...對答案感覺沒有什麼幫助就卡住了..希望高手大大幫個忙..謝謝#ed_op#/P#ed_cl#