，MMbm2Lbm

a11b11ab2121可得ABMam1bm1
αa11b11αa12b12Lαa1
b1
αabαabLαab212122222
2
αABMMMαam1bm1αam2bm2Lαam
bm

fαa11αa12Lαa1
αb11αb12Lαb1
αa21αa22Lαa2
，αBαb21αb22Lαb2
，αAMMMMMMαam1αam2Lαam
αbm1αbm2Lαbm
αa11b11αa12b12Lαa1
b1
αabαabLαab212122222
2
αAαB，MMMαam1bm1αam2bm2Lαam
bm
所以αABαAαB
fr