Advanced book on Mathematics Olympiad

Algebra 421

withak,bk∈R. It follows that

detQ(X)=

∏m

k= 1

det

[

(X+ak)^2 +b^2 kIn

]

≥ 0 ,

and the latter is positive, since for allk,

det

[

(X+ak)^2 +b^2 kIn

]

=bk^2 ndet

[(

1

bk

X+

ak

bk

) 2

+In

]

≥ 0.

In particular,Q(X)  =Yand thus the functionfis not onto.

Second solution: Because the polynomialP(t)is of even degree, the function it defines

onRis not onto. Letμbe a number that is not of the formP(t),t∈R. Then the matrix

μInis not in the image off. Indeed, ifXis ann×nmatrix, then by the spectral mapping

theorem the eigenvalues ofP(X)are of the formp(λ), whereλis an eigenvalue ofX.

Sinceμis not of this form, it cannot be an eigenvalue of a matrix in the image off. But

μis the eigenvalue ofμIn, which shows that the latter is not in the image off, and the

claim is proved.

(Gazeta Matematic ̆a(Mathematics Gazette, Bucharest), proposed by D. Andrica)

220.IfA^2 =On, then

det(A+In)=det

(

1

4

A^2 +A+In

)

=det

(

1

2

A+In

) 2

=

(

det

(

1

2

A+In

)) 2

≥ 0.

Similarly,

det(A−In)=det(−(In−A))=(− 1 )ndet(In−A)=(− 1 )ndet

(

In−A+

1

4

A^2

)

=(− 1 )ndet

(

In−

1

2

A

) 2

=(− 1 )n

(

det

(

In−

1

2

A

)) 2

≤ 0 ,

sincenis odd. Hence det(A+In)≥ 0 ≥det(A−In).

IfA^2 =In, then

0 ≤(det(A+In)^2 )=det(A+In)^2 =det(A^2 + 2 A+In)

=det( 2 A+ 2 In)= 2 ndet(A+In).

Also,

det(A−In)=(− 1 )ndet(In−A)=(− 1 )ndet

(

1

2

( 2 In− 2 A)

)