Re: rank one matrix and SVD

Janusz Kawczak wrote:
Chang wrote:

Let's say A is a column vector and B is a row vector.
Complex numbered matrix C is product of A and B.

My question is


Can we find vector A and B using singular value decomposition of C
uniquely?


Not in general and certainly not in your case! Since C has rank one, you have ONLY one non-zero eigenvalue.


Here is matlab code,


---------------------------------------------------------------------------




clear all;
A=rand(10,1)+rand(10,1)*j; % vector A is a complex column vector
B=rand(1,5)+rand(1,5)*j; % vector B is a complex row vector
A=A/norm(A);
B=B/norm(B);


C= A*B;
rank(C)


[UC, DC, VC]=svd(C);


---------------------------------------------------------------------------­-




Let's say we only have matrix C and know the rank of C is one.
I assumed the first column of UC should be same with A and the first
column of
VC is transpose of vector B.
But it is not true.


svd will only find the vectors up to a complex scaling factor.

Try dividing elements of A by elements of UC and similarly for B:
A./(UC(:,1)
B'./VC(:,1)

This will give two vectors with all constant elements, indicating that
UC = alpha*A
VC = beta*B'
where alpha and beta are complex numbers.


john2

.

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