0 0(0 0 0||0 k1|.|+k2.+…+k 得(k,k2…,kn)=0,从而k=k2=…=kn=0, 故,e2,…,en线性无关 即 得 ，从而 ， 故 线性无关． 1 2 1 0 0 0 0 1 0 0 0 0 1 0 n k k k                         + + + =                         T 1 2 ( , , , ) = 0 n k k k 1 2 0 n k k k = = = = 1 2 e e e , , , n