THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Question

Consider A M_n(C) with n 2 such that A = 0 and denote by A^* its adjutant. Show that (A^*)^2 = (tr A^*) A^*.

Accepted Answer

Since A = 0, we have rang(A) n-1. If rang(A) n-2, then A^* = O_n and (A^*)^2 = O_n (*). If rang(A) = n-1, from AA^* = ( A)I_n = O_n, results 0 rang(A) + rang(A^*) - n, therefore rang(A^*) 1. If rang(A^*) = 0, then A^* = O_n and (A^*)^2 = O_n (**). For rang(A^*) = 1, from the fact that any two lines of the matrix A^* are proportional, results A^* = BC with B = pmatrix b_1 & 0 & & 0 b_2 & 0 & & 0 & & & b_n & 0 & & 0 pmatrix, C = pmatrix c_1 & c_2 & & c_n 0 & 0 & & 0 & & & 0 & 0 & & 0 pmatrix (A^*)^2 = (BC)(BC) = B(CB)C = BDC, with D = pmatrix t & 0 & & 0 0 & 0 & & 0 & & & 0 & 0 & & 0 pmatrix, with t = _i=1^n b_i c_i = tr A^*. We get BDC = tA^*(**). The relations (*), (**), (***) are expressed through (A^*)^2 = (tr A^*) A^*.