Advanced book on Mathematics Olympiad

2.3 Linear Algebra 81

Solution.Choose a basis that identifiesVwithRmandWwithRn. Associate toAand
Btheir matrices, denoted by the same letters. The problem is solved if we prove the
equality
det(λIn−AB)=λkdet(λIm−BA),

wherekis of coursen−m. The relation being symmetric, we may assume thatn≥m.
In this case, complete the two matrices with zeros to obtain twon×nmatricesA′andB′.
Because det(λIn−A′B′)=det(λI−AB)and det(λIn−B′A′)=λn−mdet(λIn−BA),
the problem reduces to proving that det(λIn−A′B′)=det(λIn−B′A′). And this is
true for arbitraryn×nmatricesA′andB′. For a proof of this fact we refer the reader to
problem 209 in Section 2.3.2. 

IfB=A†, the transpose conjugate ofA, then this example shows thatAA†andA†A
have the same nonzero eigenvalues. The square roots of these eigenvalues are called the
singular values ofA. The second example comes from the first International Mathematics
Competition, 1994.

Example.Letαbe a nonzero real number andna positive integer. Suppose thatFand
Gare linear maps fromRnintoRnsatisfyingF◦G−G◦F=αF.

(a) Show that for allk≥1 one hasFk◦G−G◦Fk=αkFk.
(b) Show that there existsk≥1 such thatFk=On.

HereF◦GdenotesFcomposed withG.

Solution.ExpandFk◦G−G◦Fkusing a telescopic sum as follows:

Fk◦G−G◦Fk=

∑k

i= 1

(Fk−i+^1 ◦G◦Fi−^1 −Fk−i◦G◦Fi)

=

∑k

i= 1

Fk−i◦(F◦G−G◦F)◦Fi−^1

=

∑k

i= 1

Fk−i◦αF◦Fi−^1 =αkFk.

This proves (a). For (b), consider the linear mapL(F )=F◦G−G◦Facting on alln×n
matricesF. AssumingFk =Onfor allk, we deduce from (a) thatαkis an eigenvalue
ofLfor allk. This is impossible since the linear mapLacts on ann^2 -dimensional space,
so it can have at mostn^2 eigenvalues. This contradiction proves (b). 

248.LetAbea2×2 matrix with complex entries and letC(A)denote the set of 2× 2
matrices that commute withA. Prove that|det(A+B)|≥|detB|for allB∈C(A)
if and only ifA^2 =O 2.