※ 引述《mailmovieb ( )》之铭言： : 我想问说如果要证明 : A是n阶方阵，n>1 : 1.rank(A)＝n if and only if rank(adj(A))=n : 2.rank(A)＝n-1 if and only if rank(adj(A))=1 : 3.rank(A)＜n-1 if and only if adj(A)=0 : 请问该如何证明？ pf: 1. A:nonsingular <=> det(A) != 0 <=> det(adj(A)) = det(A)^(n-1) !=0 <=> adj(A): nonsingular <=> rank(adj(A)) = n 3. Suppose Aij is the cofactor of A, for any i,j, since Aij is the determinant of a n-1 * n-1 submatrix S, then rank(A) ＜ n-1 <=> S is singular, for all S <=> det(S) = 0, for all S <=> all entries of adj(A) are equal to zero <=> adj(A) = 0 <=> rank(adj(A)) = 0 2. (=>) Since det(A) = 0 => A(adj(A)) = 0 => R(adj(A)) is contained in N(A) => rank(adj(A)) ≦ nullity(A) = n - rank(A) = 1 since rank(A)=n-1, applying 3 we can get rank(adj(A)) != 0, therefore rank(adj(A)) = 1 (<=) we can know that rank(A) ≧ n-1 but is not equal to n (by 1 and 3), rank(A) can only be n-1. --

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