Question

设A、B均为正交矩阵,且的A绝对值等于B的负的绝对值,实证和的绝对值等于0

Answer 1

证明:由A、B均为正交矩阵

所以 A^TA = AA^T = E

B^TB = BB^T = E

又因为 |A||B| = -1

所以 -|A+B|

= |A||A+B||B|

= |A^T||A+B||B^T|

= |A^T(A+B)B^T|

= |A^TAB^T+A^TBB^T|

= |B^T+A^T|

= |(B+A)^T|

= |A+B|

所以有 2|A+B| = 0

所以 |A+B| = 0.