ronnor
幼苗
共回答了17个问题采纳率:94.1% 举报
m*n矩阵A的秩为 r( r>=1 ) 存在可逆矩阵P,Q使得
PAQ=diag(1,1,...1,0...)(共r个1,这就是A的标准型)
A=P^(-1)diag(1,1,...1,0...)Q^(-1) =P^(-1)diag(1,0,0...0,0...0)Q^(-1)+P^(-1)diag(0,1,0,...0,0...0)Q^(-1)+P^(-1)diag(0,0,1,...0,0...0)Q^(-1)+...+P^(-1)diag(0,0,0...1,0...0)Q^(-1).
又diag(1,0,0...0,0...0)=(1,0,0...0,0...0)'(1,0,0...0,0...0),
P^(-1)diag(1,0,0...0,0...0)Q^(-1)=
P^(-1)(1,0,0...0,0...0)'(1,0,0...0,0...0)Q^(-1)
设P^(-1)(1,0,0...0,0...0)'=a1,(1,0,0...0,0...0)Q^(-1)=b1',都是列向量!
同理可以构造出a2,...ar与b2,.br,显然a1,...,ar与b1,...,br为线性无关的向量组
且A=a1b1'+a2b2'+...+arbr'=∑aibi'(i从1到r)
1年前
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