又N(A)gN(AA)故N(A)=N(AA) 在R(AA)=R(A)中,将A换为,A换为(A),则有 RQ(A2)=R((A)2)=R(AA)) (7)以AA= I e rankA=m为例 rankA=rank(AA)=rank(I)=m rank(Aa)=rankA=m 即AA为m阶满秩可逆方阵,(AA)存在。 又AA幂等:(AA0)=AA,乘以(A4),得AA"=ln (8)R(A)={Ax|x∈"}cC R(AB)={ABy|y∈C"}sCm→R(A2R(AB) 对AB(AB)A=A台rmk(AB)= ranke(A ranke4=rmk(AB(AB)A)≤rwmk(AB)≤rmk(A) rank(AB)=rankA rankA = dim R(A), rank(AB)=dim R(aB) 故R(A)=R(AB) 即,Ⅴx∈C",y∈C,使Ax=ABy.故 Ax=ABy= AB(AB)ABy= AB(AB)Ax (注意x∈C")→AB(AB)"A=A Xf B(AB)AB=B rank(AB)=rank(B) rankB=rumk(B(AB)AB)≤rwk(AB)≤rmk(B) → rank(AB)= rankB ∈R(B(AB)AB)={B(AB)ABy|y∈C"}≤R(B)={Bx|x∈Cn} 又, rankB=rmhk(AB)=rmk(AB(AB)AB)≤rmhk(B(AB)AB)≤rmkB →rmkB=rmk(B(AB)AB)→R(B)=R(B(AB)yAB)又 (1) N A N A A ( ) ( )  故 (1) N A N A A ( ) ( ) = 在 (1) R AA R A ( ) ( ) = 中，将 A 换为 H A ， (1) A 换为 (1) ( )H A ,则有 (1) (1) ( ) ( ( ) ) (( ) ) H H H H R A R A A R A A = = (7) 以 (1) AA I rankA m =  = m 为例. (1) : ( ) ( ) . m  = = = rankA rank AA rank I m (1)  = = : ( ) rank AA rankA m 即 (1) AA 为 m 阶满秩可逆方阵, (1) 1 ( ) AA − 存在。 又 (1) AA 幂等： (1) 2 (1) （AA AA ）＝ , 乘以 (1) 1 AA （ ）－ ，得 (1) AA I ＝ m (8) ( ) { | }n m R A Ax x C C =   ( ) { | } ( ) ( ) P m R AB ABy y C C R A R AB =   →  (1) •对 ( ) ( ) ( ) AB AB A A rank AB rank A =  = (1) : ( ( ) ) ( ) ( ) ( ) rankA rank AB AB A rank AB rank A rank AB rankA  =   → = : dim ( ), ( ) dim ( ) ( ) ( ) rankA R A rank AB R AB R A R AB  = = 故 = 即， , , n p     x C y C 使 Ax ABy = . 故 (1) (1) (1) ( ) ( ) ( ) ( ) n Ax ABy AB AB ABy AB AB Ax x C AB AB A A = = = 注意  → = • 对 (1) B AB AB B rank AB rank B ( ) ( ) ( ) =  = (1) : ( ( ) ) ( ) ( ) ( ) rankB rank B AB AB rank AB rank B rank AB rankB  =   → = (1) (1) : ( ( ) ) { ( ) | } ( ) { | } p p  =   =  R B AB AB B AB ABy y C R B Bx x C 又， (1) (1) rankB rank AB rank AB AB AB rank B AB AB rankB = =   ( ) ( ( ) ) ( ( ) ) (1) (1) → =  = rankB rank B AB AB R B R B AB AB ( ( ) ) ( ) ( ( ) )