矩阵A,B,且AB=BA,怎么证明(A+B)^n=C(n,0)A^n+C(n,1)A^(n-1)B+C(n,2)A^(n-2)B^2+...+C(n,n)B^n

Question

矩阵A,B,且AB=BA ,怎么证明(A+B)^n = C(n,0)A^n+C(n,1)A^(n-1)B+C(n,2)A^(n-2)B^2+...+C(n,n)B^n

Answer 1

用数学归纳法
n=1时,显然有 (A+B)^1 = C(1,0)A+ C(1,1)B
设n=k时,成立,即 (A+B)^k = C(k,0)A^k+C(k,1)A^(k-1)B+C(k,2)A^(k-2)B^2+...+C(k,k)B^k
n=k+1时
(A+B)^(k+1)=(A+B)^k (A+B) = (C(k,0)A^k+C(k,1)A^(k-1)B+C(k,2)A^(k-2)B^2+...+C(k,k)B^k)(A+B)
=C(k,0)A^(k+1)+(C(k,1)+C(k,0))A^kB+C(k,2)A^(k-2)B³+...+ (C(k,k)+C(k,k-1))AB^k + C(k,k)B^(k+1)
=C(k+1,0)A^(k+1)+C(k+1,1)A^kB+...+C(k+1,k)AB^k + C(k+1,k+1)B^(k+1)
也成立.
故成立.交换性用在相乘的时候有 A^iB^j · A = A^(i+1)B^j