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IMC / 1995 / Problems / Day 1, P5

IMC 1995 · Day 1 · P5

Let AA and BB be real n×nn \times n matrices. Assume that there exist n+1n+1 different real numbers t1,t2,,tn+1t_1, t_2, \dots, t_{n+1} such that the matrices Ci=A+tiB,i=1,2,,n+1,C_i = A + t_i B, \quad i = 1, 2, \dots, n+1, are nilpotent (i.e. Cin=0C_i^n = 0).

Show that both AA and BB are nilpotent.

Solution (official)

We have that (A+tB)n=An+tP1+t2P2++tn1Pn1+tnBn(A + tB)^n = A^n + t P_1 + t^2 P_2 + \cdots + t^{n-1} P_{n-1} + t^n B^n for some matrices P1,P2,,Pn1P_1, P_2, \dots, P_{n-1} not depending on tt.

Assume that a,p1,p2,,pn1,ba, p_1, p_2, \dots, p_{n-1}, b are the (i,j)(i,j)-th entries of the corresponding matrices An,P1,P2,,Pn1,BnA^n, P_1, P_2, \dots, P_{n-1}, B^n. Then the polynomial btn+pn1tn1++p2t2+p1t+ab t^n + p_{n-1} t^{n-1} + \cdots + p_2 t^2 + p_1 t + a has at least n+1n+1 roots t1,t2,,tn+1t_1, t_2, \dots, t_{n+1}. Hence all its coefficients vanish. Therefore An=0A^n = 0, Bn=0B^n = 0, Pi=0P_i = 0; and AA and BB are nilpotent.

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