M=a bT
A.det(M)
wlog M's eigenvalue k and eigenvector v
Mv=kv => bTMv=kbTv => bTa(bTv)=k(bTv)
(1)if bTv≠0,then k=bTa.
(2)if bTv=0,then k=0.
Mv=kv => a(bTv)=kv => k=0 since v≠0.
Hence M have eigenvalue k=0 and bTa
det(M)=k_1*k_2*......*k_n=0
B.rank(M)=1
M=[c_1 c_2 ...... c_n],where c_i=[a_1*b_i a_2*b_i ... ... a_n*b_i] i=1,2,......,n ,
rank(M)=dim(span{ c_1 , c_2 , ...... , c_n })=dim(span{c_1})=1
C.trace(M)=bTa
rank(M)=1 , dimension theorem nullity(M)=n-1,
that is;exist v_1,v_2,...,v_(n-1) such that N(M)=span{v_1,v_2,......,v_(n-1)},
N(M)=N(M-0*I),so 0 is a root of multiolicity n-1 of det(M-x*I) and bTa is a simple root,
trace(M)=0+0+......+0+bTa=bTa