Advanced book on Mathematics Olympiad

Algebra 445

LetPB(λ)=λ^2 +rλ+sbe the characteristic polynomial ofB. By the Cayley–Hamilton
Theorem,PB(B)=0. We have

O 2 =APB(B)−PB(B)A=AB^2 −B^2 A+r(AB−BA)= 2 BC+rC.

Using this and the fact thatCcommutes withAandB, we obtain

O 2 =A( 2 BC+rC)−( 2 BC+rC)A= 2 (AB−BA)C= 2 C^2.

Therefore,C^2 =O 2. In some basis

C=

(

0 α

00

)

.

Hence C commutes only with polynomials inC. But ifAandBare polynomials inC,
thenC=O 2 , a contradiction. SoCmust be scalar whose square is equal to zero, whence
C=O 2 again. This shows that such matricesAandBdo not exist.
(American Mathematical Monthly, solution by W. Gustafson)

263.Chooseλ∈Rsufficiently large such thatλIn+Ahas positive entries. By the
Perron–Frobenius Theorem, the largest eigenvalueρofλIn+Ais positive, and all other
eigenvalues lie inside the circle of radiusρcentered at the origin. In particular,ρis real
and all other eigenvalues lie strictly to its left. The eigenvalues ofAare the horizontal
translates byλof the eigenvalues ofλIn+A, so they enjoy the same property.

Remark.The result is true even for matrices whose off-diagonal entries are nonnegative,
the so-called Metzler matrices, where a more general form of the Perron–Frobenius
Theorem needs to be applied.

264.First solution: DefineA=(aij)^3 i,j= 1. Then replaceAbyB =αI 3 −A, where
αis chosen large enough so that the entriesbij of the matrixBare all positive. By
the Perron–Frobenius Theorem, there exist a positive eigenvalueλand an eigenvector
c=(c 1 ,c 2 ,c 3 )with positive coordinates. The equalityBc=λcyields

a 11 c 1 +a 12 c 2 +a 13 c 3 =(α−λ)c 1 ,

a 21 c 1 +a 22 c 2 +a 23 c 3 =(α−λ)c 2 ,

a 31 c 1 +a 32 c 2 +a 33 c 3 =(α−λ)c 3.

The three expressions from the statement have the same sign asα−λ: they are either all
three positive, all three zero, or all three negative.

Second solution: The authors of this problem had a geometric argument in mind. Here
it is.