# [高等代数] 一道关于半正定的高代题目求教

 已知$A,B$是同阶半正定，$Tr(AB)=0,$ 则 $（A+B）^m=A^m+B^m.$

 $A, B$is positive semidefinite, then $AB$ and $BA$ are also positive semidefinite.         Because of $tr (BA) =tr (AB) =0$,hence all eigenvalues of $AB, BA$are 0, this is, $AB=BA=0$. Using binomial theorem have $(A+B)^m=A^m+B^m$.

GMT+8, 2019-4-19 00:35 , Processed in 1.046875 second(s), 16 queries , Gzip On.