由定义,正交正定矩阵a,a*a'=a'*a=E;
另外显然有a*E=E*a=E;
比较二式,由于ab=ba=E中如果a、b正定,对正定的a,有b唯一,(正定的b,有a唯一),
所以b=E,同理证得a=E;所以,a=b=a'=b’=E .
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