[大學]線性代數一問?

[大學]線性代數一問?

MasterBow 於 星期日 九月 06, 2009 9:44 am


題目很簡單~但是達案跟我做的不一樣........
題目是這樣的
註(bT 是 b得反置矩証) (Rn 有N個元素)
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.
answer is 0
我目前的做法是
因為定理有一敘述是 det(A +or - B) =det(A) +or -det (B)
因此題目鎖有matric加總變成
  45 45 45
[ 45 45 45 ] ---->det()=0
  45  45 45
但是有但書 det(A +or - B) =det(A) +or -det (B) =>(不是所有狀況都成立)
這樣做有風險 因為定裡不是絕對?
有其他其他方式? 或者證明可以使用det(A +or - B) =det(A) +or -det (B) ??

感激不盡!!

MasterBow
初學者
初學者
 
文章: 1
註冊時間: 2009-09-06

danny 於 星期二 七月 23, 2013 4:38 am


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
實習生
實習生
 
文章: 64
註冊時間: 2005-08-07






代數學