Advanced book on Mathematics Olympiad

420 Algebra

and the latter is nonnegative, as seen in the introduction. Taking the limit withap-
proaching zero, we obtain det(A^2 m+B^2 m)≥0.
Forkodd,k= 2 m+1, letx 0 =− 1 ,x 1 ,x 2 ,...,x 2 mbe the zeros of the polynomial
x^2 m+^1 +1, withxj+m=xj,j= 1 , 2 ,...,m. BecauseAandBcommute, we have

A^2 m+^1 +B^2 m+^1 =(A+B)

∏m

j= 1

(A−xjB)(A−xjB).

SinceAandBhave real entries, by taking determinants we obtain

det(A−xjB)(A−xjB)=det(A−xjB)det(A−xjB)

=det(A−xjB)det(A−xjB)

=det(A−xjB)det(A−xjB)≥ 0 ,

forj= 1 , 2 ,...,m. This shows that the sign of det(A^2 m+^1 +B^2 m+^1 )is the same as the
sign of det(A+B)and we are done.
(Romanian Mathematical Olympiad, 1986)

218.The caseλ≥0 was discussed before. Ifλ<0, letω=

√

−λ. We have

det(In+λA^2 )=det(In−ω^2 A^2 )=det(In−ωA)(In+ωA)

=det(In−ωA)det(In+ωA).

Because−A=At, it follows that

In−ωA=In+ωAt=t(In+ωA).

Therefore,

det(In+λA^2 )=det(In+ωA)dett(In+ωA)=(det(In+ωA))^2 ≥ 0 ,

and the inequality is proved.
(Romanian mathematics competition, proposed by S. R ̆adulescu)

219.First solution: We can assume that the leading coefficient ofP(t)is 1. Letα
be a real number such thatP(t)+αis strictly positive and letY be a matrix with
negative determinant. Assume thatfis onto. Then there exists a matrixXsuch that
P(X)=Y−αIn.
Because the polynomialQ(t)=P(t)+αhas no real zeros, it factors as

Q(t)=

∏m

k= 1

[

(t+ak)^2 +b^2 k

]