Advanced book on Mathematics Olympiad

(ff) #1

412 Algebra


Remark.This decomposition plays a special role, especially for linear operators on
infinite-dimensional spaces. IfAandBcommute, thenMis called normal.


200.The answer is negative. The trace ofAB−BAis zero, while the trace ofInisn;
the matrices cannot be equal.


Remark.The equality cannot hold even for continuous linear transformations on an
infinite-dimensional vector space. IfP andQare the linear maps that describe the
momentum and the position in Heisenberg’s matrix model of quantum mechanics, and if
is Planck’s constant, then the equalityPQ−QP=Iis the mathematical expression
of Heisenberg’s uncertainty principle. We now see that the position and the momentum
cannot be modeled using finite-dimensional matrices (not even infinite-dimensional con-
tinuous linear transformations). Note on the other hand that the matrices whose entries
are residue classes inZ 4 ,


A=





0100

0010

0001

0000




⎠ and B=





0000

1000

0200

0030




⎠,

satisfyAB−BA=I 4.


201.To simplify our work, we note that in general, for any two square matricesAandB
of arbitrary dimension, the trace ofAB−BAis zero. We can therefore write


AB−BA=

(

ab
c−a

)

.

But then(AB−BA)^2 =kI 2 , wherek=a^2 +bc. This immediately shows that an odd
power ofAB−BAis equal to a multiple of this matrix. The odd power cannot equalI 2
since it has trace zero. Therefore,nis even.
The condition from the statement implies thatkis a root of unity. But there are only
two real roots of unity and these are 1 and−1. The squares of both are equal to 1. It
follows that(AB−BA)^4 =k^2 I 2 =I 2 , and the problem is solved.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu)


202.Assume thatp =q. The second relation yieldsA^2 B^2 =B^2 A^2 =rA^4 andrB^2 A=
rAB^2 =A^3. Multiplying the relationpAB+qBA=Inon the right and then on the
left byB, we obtain


pBAB−qB^2 A=B and pAB^2 +qBAB=B.

From these two identities and the fact thatB^2 A=AB^2 andp =qwe deduceBAB=
AB^2 =B^2 A. Therefore,(p+q)AB^2 =(p+q)B^2 A=B. This implies right away that

Free download pdf