证明：
首先，|a|^2=a'a
设 Pa=b
|b|^2=b'b=(Pa)'(Pa)=aP'Pa=a(P'P)a=a'a,(根据正交矩阵性质，P'P=I)
所以，|a|=|b|.
abilitytao 2013-02-23 18:48 发表评论
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