【公式证明】矩阵乘积转置

证明：
(AB)T=BTAT(AB)^T=B^TA^T(AB)T=BTAT
对于等式左侧
设AB=C,CT=DAB=C,C^T=DAB=C,CT=D
则Dm,n=Cn,m=[An,:,B:,m]D_{m,n}=C_{n,m}=[A_{n,: },B_{:,m}]Dm,n​=Cn,m​=[An,:​,B:,m​]
对于等式右侧
设BTAT=GB^TA^T=GBTAT=G
则Gm,n=[Bm,:T,A:,nT]G_{m,n}=[B^T_{m,:},A^T_{:,n}]Gm,n​=[Bm,:T​,A:,nT​]
由于Bm,:T=B:,m,A:,nT=An,:B^T_{m,:}=B_{:,m},A^T_{:,n}=A_{n,:}Bm,:T​=B:,m​,A:,nT​=An,:​
所以Gm,n=[B:,m,An,:]G_{m,n}=[B_{:,m},A_{n,: }]Gm,n​=[B:,m​,An,:​]
又由于[An,:,B:,m]=[B:,m,An,:][A_{n,: },B_{:,m}]=[B_{:,m},A_{n,: }][An,:​,B:,m​]=[B:,m​,An,:​]
得出Dm,n=Gm,nD_{m,n}=G_{m,n}Dm,n​=Gm,n​
因此(AB)T=BTAT(AB)^T=B^TA^T(AB)T=BTAT