1.
设AX=β有解.
任意Y,A^tY=0
==><Y,β>=<Y,AX>=<A^tY,X>=<0,X>=0
2.
记KerA^t={Y,A^tY=0},ImA={AX,任意X}
设β∈[KerA^t]^┴
==>
Dim{[KerA^t]^┴}=n-Dim[KerA^t]=n-[n-R(A^t)]=
=R(A^t)=R(A)=Dim[ImA]
任意AX∈ImA,任意Y∈KerA^t
<Y,AX>=<A^tY,X>=<0,X>=0
==>AX∈[KerA^t]^┴
==>
ImA为[KerA^t]^┴的子空间,而它们维数相同,所以
ImA=[KerA^t]^┴
==>有X,AX=β.
