[大學]線性代數一問?

[大學]線性代數一問?

let m=a(bT) where a and b are two nonzero column vector of (Rn) Please answer the follow question
(A)what is  rank(M)? det(M)? and trace(M)?
I think....
if a =[a1]       [b1]         so Rank(M)=Rank(abT) is N
a2         b2
..      b= ..
..           ..
an         bn
but the absolute answer is Rank(M)=1...? Why?
therefore det(M) trace(M) also error...

...............................................................................................
Digit 1 to 9 can be arranded into 3-3 matrices in 9! ways. Find the sum of the
determinant of these matrices.

45 45 45
[ 45 45 45 ] ---->det()=0
45  45 45

MasterBow

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

danny