Problem 12173
(American Mathematical Monthly, Vol.127, March 2020) Proposed by F. Stanescu (Romania).
Suppose that X and Y are n-by-n complex matrices such that 2Y2 = XY − Y X and the rank of X − Y is 1. Prove Y3= Y XY .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let Z = (Y − X)/2 then Z has rank 1 and there exist v, w ∈ Cn\ {0} such that Z = vwT. Moreover, since X = Y − 2Z, it follows that
2Y2= XY − Y X ⇔ Y2= Y Z − ZY and Y3= Y XY ⇔ Y ZY = 0.
Hence we have to prove that Y ZY = 0.
We first show that the eigenvalues of Y are all zero that is Y is nilpotent.
Let λ1, . . . , λr the pairwise distinct eigenvalues of Y and m1, . . . , mr their multiplicities. For any integer k ≥ 0
Y2+k= YkY Z − YkZY = Y (YkZ) − (YkZ)Y and therefore
Xr i=1
tiλki = Xr i=1
miλ2+ki = tr(Y2+k) = tr(Y (YkZ)) − tr((YkZ)Y ) = 0
where ti = miλ2i for i = 1, . . . , r. The (Vandermonde) determinant related to this homogeneous system of linear equations in the unknowns t1, . . . , tr is nonzero because λ1, . . . , λr are pairwise distinct. Hence, it follows that t1= · · · = tr= 0, that is r = 1 and λ1= 0.
Finally we verify that Y ZY = 0. We have two cases.
1) If v is an eigenvector of Y then Y v = 0 and
Y ZY = Y (vwT)Y = (Y v)(wTY ) = 0.
2) If v is not an eigenvector of Y then v and Y v are linearly independent and the 2-dimensional vector space V generated by v and Y v is invariant with respect to Y :
Y v ∈ V and Y (Y v) = Y2v = Y Zv − ZY v = (wTv)Y v − (wTY v)v ∈ V.
Since Y is nilpotent and V has dimension 2, by Cayley-Hamilton theorem, it follows that the restriction of Y2 to V is identically zero and therefore
0 = Y2v = (wTv)Y v − (wTY v)v =⇒ (wTv) = 0 and (wTY v) = 0.
In particular Z2= (vwT)(vwT) = v(wTv)wT = 0. Therefore
0 = (Z + Y )(Y2− (Y Z − ZY )) + (Y2− (Y Z − ZY ))(Z − Y ) = 2Y ZY + Z2Y − Y Z2= 2Y ZY + 0 + 0
which implies Y ZY = 0 and we are done.
