aika
设A是实对称矩阵,且A^2=0,证明:A=0
A是实对称矩阵 A=﹙aij﹚ aij=aji 从aij=0 可得aij=0看A的i行i列交点元素﹙A﹚ii=∑[1≤k≤n]aikaki=∑[1≤k≤n]aikaik=∑[1≤k≤n]aik=0[∵A=0]∴aik=0 aik=0 A的第i行全为0.i任意.A的每一行都全为0.A=0 解析看不懂?免费查看同类题视
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