定理1设A,B∈C,A为正定矩阵,BB=B 则存在可逆矩阵T,使得 T"AT=E, T BT=D 证:A正定→A与E合同→PHAP=E →PBBP为 Hermite矩阵 今從pBPU=D T=PU BT=D T=PU THAT= UHPHAPUEUHEU=E定理1 , , , , n n H A B C A B B T  设 为正定矩阵,  = 则存在可逆矩阵 使得 证: 正定 与 合同 A A E  , . H H T AT E T BT D = = n H  = P AP E H  P BP Hermite 为 矩阵 H H  = U P BPU D H T PU T BT D = = H H H H T PU T AT U P APU U EU E = = = =