A^2=A得到A(A-E)=0由r(A)+r(B)-n<=r(AB)所以r(A)+r(A-E)-n<=r(A(A-E))=0所以r(A)+r(A-E)<=n有由于r(A)+r(B)>=r(A±B)所以r(A)+r(A-E)>=r[A-(A-E)]=r(E)=n所以n<=r(A)+r(A-E)<=n所以r(A)+r(A-E)=n