具体一点，谢谢了
用最基本的方法：设A==(a ij)m*n 分块A==(A1,A2,...,An),Aj==(a 1j,a 2j,...,a mj)(j==1,2,...n)
则T(A)==T(T(A1),T(A2),...,T(An))
∴AT(A)==∑AjT(Aj)(j==1,2,...n) 显然Aj为袭做m*1阵T(Aj)为1*m阵 故AT(A)必为m*m阵
考虑乘积矩阵尺首对角线的元有(a 1j)^2==(a 2j)^2==...==(a mj)^2==0
故a 1j==a 2j==...==a mj==0.又j==1,2,...n
∴a ij==0,i==1,2...,m,j==1,2,...n
即A==O 得陵禅数证
A*A'=B
=>/A/ * /A'/ = /B/ =0
=> /A/ * /备链塌A/ =0
=> /唤凳A/ =0
秩为零 我只能仿圆证出这点
不清楚啊！！！！