【公式证明】矩阵乘积转置
证明：
( A B ) T = B T A T (AB)^T=B^TA^T (AB)T=BTAT
对于等式左侧
设 A B = C , C T = D AB=C,C^T=D AB=C,CT=D
则 D m , n = C n , m = [ A n , : , 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 B T A T = G B^TA^T=G BTAT=G
则 G m , n = [ B m , : T , A : , n T ] G_{m,n}=[B^T_{m,:},A^T_{:,n}] Gm,n=[Bm,:T,A:,nT]
由于 B m , : T = B : , m , A : , n T = A n , : B^T_{m,:}=B_{:,m},A^T_{:,n}=A_{n,:} Bm,:T=B:,m,A:,nT=An,:
所以 G m , n = [ B : , m , A n , : ] G_{m,n}=[B_{:,m},A_{n,: }] Gm,n=[B:,m,An,:]
又由于 [ A n , : , B : , m ] = [ B : , m , A n , : ] [A_{n,: },B_{:,m}]=[B_{:,m},A_{n,: }] [An,:,B:,m]=[B:,m,An,:]
得出 D m , n = G m , n D_{m,n}=G_{m,n} Dm,n=Gm,n
因此 ( A B ) T = B T A T (AB)^T=B^TA^T (AB)T=BTAT