(1)∵V∈W,有B=∑ka1 y∈L( m+1 n),有 于是(,B)=∑k∑1(a,a1)=0 j=m+1 (2)任=∑xa1∈V,若以⊥W,则a,a,)=0 x=(x1,a,)=(a,ax)=0, j=1,2,…,m. a=∑xa1∈L(am+1,,an) i=m+110 i m i i  W  k =   = 1 (1) , 有 2 , , ( , ) 0, 1 =   ⊥ = = j n i （ ） 任 xi  i V 若 W 则   j = 1,2,  ,m j = 1,2,  ,m. ( , ) ( , ) 0 ( , , ), 1 1 1 1 = =   =    = = + = + + n j m j i j m i i n j m m n j j k L            于是  有 ( , ) ( , ) 0, 1  =  = = = j j n i x j xi  i    ( , , ). 1 1 m n n i m  xi  i L  +   = +  =  