Advanced book on Mathematics Olympiad

440 Algebra

the kernel, in which case their eigenvalue is 0. BecauseA= 2 B−I, it has the same

eigenvectors asB, with eigenvalues±1. This proves (a).

Part (b) is based on the fact that any family of commuting diagonalizable operators

onVcan be diagonalized simultaneously. Let us prove this property by induction on the

dimension ofV. If all operators are multiples of the identity, there is nothing to prove.

If one of them, sayS, is not a multiple of the identity, then consider the eigenspaceVλ

of a certain eigenvalueλ.IfT is another operator in the family, then sinceST v =

TSv=λT v, it follows thatTv∈Vλ; henceVλis an invariant subspace for all operators

in the family. This is true for all eigenspaces ofA, and so all operators in the family

are diagonal blocks on the direct decomposition ofVinto eigenvectors ofA. By the

induction hypothesis, the family can be simultaneously diagonalized on each of these

subspaces, and so it can be diagonalized on the entire spaceV.

Returning to the problem, diagonalize the pairwise commuting involutions. Their

diagonal entries may equal+1or−1 only, showing that there are at most 2nsuch

involutions. The number can be attained by considering all choices of sign on the diagonal.

(3rd International Competition in Mathematics for University Students, 1996)

254.From the orthogonality ofAuandu, we obtain

〈Au, u〉=〈u, Atu〉=〈Atu, u〉= 0.

Adding, we obtain that〈(A+At)u, u〉=0 for every vectoru. ButA+Atis symmetric,
hence diagonalizable. For an eigenvectorvof eigenvalueλ, we have

〈(A+At)v, v〉=〈λv, v〉=λ〈v, v〉= 0.

This shows that all eigenvalues are zero, soA+At=0, which proves (a).

As a corollary of this, we obtain thatAis of the form

A=

⎛

⎝

0 a 12 a 13

−a 12 0 a 23

−a 13 −a 23 0

⎞

⎠.

SoAdepends on only three parameters, which shows that the matrix can be identified

with a three-dimensional vector. To choose this vector, we compute

Au=

⎛

⎝

0 a 12 a 13

−a 12 0 a 23

−a 13 −a 23 0

⎞

⎠

⎛

⎝

u 1

u 2

u 3

⎞

⎠=

⎛

⎝

a 12 u 1 +a 13 u 2

−a 12 u 1 +a 23 u 3

−a 13 u 1 −a 23 u 2

⎞

⎠.

It is easy to see now that if we setv=(−a 23 ,a 13 ,−a 12 ), thenAu=v×u.

Remark.The set of such matrices is the Lie algebra so( 3 ), and the problem describes two

of its well-known properties.