Problem 10930
(American Mathematical Monthly, Vol.109, March 2002) Proposed by P. Schweitzer (USA).
Show that there do not exist real 2-by-2 matices A and B such that their commutator is nonzero and commutes with both A and B. (The commutator C of A and B is defined by C = AB − BA.) Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By the Cayley-Hamilton theorem, the matrices A and B solve their characteristic polynomial:
A2= aA + b , B2= cB + d
where a = tr(A), b = −det(A) and c = tr(B), d = −det(B). Since C commutes with A
AC − CA= 0
AC+ CA = A2B+ BA2= (aA + b)B − B(aA + b) = aC Therefore, AC = CA = a2Cand, in the same way, BC = CB = c2C. Thus
C2= (AB − BA)C = A(BC) − B(AC) = a 2 ·c
2C −c 2 ·a
2C= 0.
Assume that C 6= 0. Then there is a vector v 6= 0 such that u := Cv 6= 0. Note that u is a common eigenvector of C, A and B:
Cu= C2v= 0, Au = ACv = a
2Cv= a
2u, Bu= BCv = c
2Cv= c 2u.
The vectors u and v are linearly independent: let α, β ∈ R such that αu + βv = 0 then 0 = C(αu + βv) = αCu + βCv = αC2v+ βCv = βu,
which implies that α = β = 0. Since a = tr(A) and c = tr(B) then the representations of the matrices A and B with respect to the base < u, v > are
A=
a
2 γ
0 a2
and B =
c
2 δ
0 2c
for some γ, δ ∈ R. Then
C= AB − BA =
ac 4
a 2δ+2cγ 0 ac4
−
ac 4
c 2γ+a2δ 0 ac4
= 0
that is a contradiction.
Remark. The statement does not hold in higher dimension: for example in R3×3
if A =
0 1 0 0 0 0 0 0 0
and B =
0 0 0 0 0 1 0 0 0
then C = AB − BA =
0 0 1 0 0 0 0 0 0
and AC = CA, BC = CB.
