Question

高等代数求交空间的基

β2=(1,-1,3,7)

Answer 1

L(a1T,a2T)+L(b1T,b2T)=L(a1T,a2T,b1T,b2T)
对(a1T,a2T,a3T,a4T)
=
1 -1 2 1
2 1 -1 -1
1 1 0 3
0 1 1 7
做初等行变换
~r2-2r1,r3-r1
1 -1 2 1
0 3 -5 -3
0 2 -2 2
0 1 1 7
~r3*1/2,r2-3r3,r4-r3
1 -1 2 1
0 0 -2 -6
0 1 -1 1
0 0 2 6
~r2<->r3,r3*(-1/2),r4-2r3
1 -1 2 1
0 1 -1 1
0 0 1 3
0 0 0 0
~r1+r2,r2+r3
1 0 1 2
0 1 0 4
0 0 1 3
0 0 0 0
~r1-r3
1 0 0 -1
0 1 04
0 0 1 3
0 0 0 0
容易看出，dimL(a1,a2)∪L(b1,b2)=3,{a1,a2,b1}是一组基，还有b2=-a1+4a2+3b1
因为dimL(a1,a2)∩dimL(b1,b2)=dimL(a1,a2)+dim(b1,b2)-dimL(a1,a2)∪L(b1,b2)
=2+2-3=1,{a1-4a2}是基。