438 Algebra
It follows thatλ 1 =λ 2 =0. Change the basis tov, wwithvan eigenvector ofA(which
does exist becauseAv=0 has nontrivial solutions). This transforms the matrix into one
of the form
(
0 a
00
)
.
One easily checks that the square of this matrix is zero.
Conversely, assume thatA^2 =O 2. By the spectral mapping theorem both eigenvalues
ofAare zero, so by appropriately choosing the basis we can makeAlook like
(
0 a
00
)
.
Ifa=0, we are done. If not, then
C(A)=
{(
αβ
0 α)
|α, β∈R}
.
One verifies immediately that for everyB∈C(A), det(A+B)=detB. So the inequality
from the statement is satisfied with equality. This completes the solution.
(Romanian Mathematical Olympiad, 1999, proposed by D. Mihe ̧t)
249.Since detB=1,Bis invertible andB−^1 has integer entries. From
A^3 +B^3 =((AB−^1 )^3 +I 2 )B^3 ,it follows that det((AB−^1 )^3 +I 2 )=1. We will show that(AB−^1 )^2 =O 2. SetAB−^1 =C.
We know that det(C^3 +I 2 )=1. We have the factorization
C^3 +I 2 =(C+I 2 )(C+I 2 )(C+^2 I 2 ),whereis a primitive cubic root. Taking determinants, we obtain
P(− 1 )P (−)P(−^2 )= 1 ,wherePis the characteristic polynomial ofC.
LetP(x)=x^2 −mx+n; clearlym, nare integers. BecauseP(−^2 )=P(−)=
P(), it follows thatP(−)P(−^2 )is a positive integer. SoP(− 1 )=P(−)P(−^2 )=
- We obtain 1+m+n=1 and(^2 +m+n)(+m^2 + 1 )=1, which, after some
algebra, givem=n=0. SoChas just the eigenvalue 0, and being a 2×2 matrix, its
square is zero.
Finally, from the fact thatAB=BAand(AB−^1 )^2 =O 2 , we obtainA^2 B−^2 =O 2 ,
and multiplying on the right byB^2 we haveA^2 =O 2 , as desired.
(Romanian Mathematics Competition, 2004, proposed by M. Becheanu)