设向量a=(1 1 1 ... 1)T显然J=(1 1 1 ... 1)T(1 1 1 ... 1)=aaTJJ=（aaT）（aaT）=a(aTa)aT = a(n)aT = n(aaT)=nJ则（E-J）（E-J/(n-1)）=E-J-J/(n-1)+JJ/(n-1)=E-J-J/(n-1)+nJ/(n-1)=E因此E-J可逆，且逆矩阵是E-J/(n-1)
