设A是m*n的矩阵,证明若对任意m维行向量x和n维列向量,都有xAy=o,则A=0
人气:473 ℃ 时间:2020-01-30 06:30:13
解答
证明: 设 A = (aij).
取xi 是第i个分量为1其余分量为0的m维行向量, i=1,2,…,m;
取yj是第j个分量为1其余分量为0的n维列向量, j=1,2,…,n.
则有 xi A yj = aij,i=1,2,…,m; j=1,2,…,n .
若对任意m维行向量x和n维列向量,都有xAy=o, 则必有
xi A yj = aij = 0,i=1,2,…,m; j=1,2,…,n
故有 A = 0.
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