Representing a matrix as a product of two matrices in MATLAB

Question

rihab 2015-11-17

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评论： Torsten 2015-11-19

I have a 4x4 matrix of complex numbers, say

X = [0.4079 + 0.0000i   0.7532 + 0.0030i   0.9791 + 0.0272i   0.9335 - 0.0036i;
7532 - 0.0030i   1.2288 + 0.0000i   1.3074 + 0.0052i   0.9791 + 0.0272i;
9791 - 0.0272i   1.3074 - 0.0052i   1.2288 + 0.0000i   0.7532 + 0.0030i;
9335 + 0.0036i   0.9791 - 0.0272i   0.7532 - 0.0030i   0.4079 + 0.0000i]

I want to represent it as a product of a 4x1 & 1x4 vectors say x such that X = xx^H where H denotes hermitian transpose of x. does anyone have any hint/suggestion to solve this matrix decomposition problem? Any suggestion would be appreciated.

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rihab 2015-11-18

I see. I used SVD to decompose this matrix X. However, the matrix UV^H (i obtained U & V after using svd command of MATLAB) doesnt have rank 1. Could you be a bit elaborate on what you meant by "It could be done with a rank 1 matrix"? I would appreciate suggestions.

rihab 2015-11-18

@stefan the matrix X is a solution to a least squares problem (AX-B). X is a 16x1 vector and for my purpose I have reshaped into a 4x4 matrix and now I want to decompose this matrix X as a product of two vectors such that X = xx^H (I am basically interested in extracting the four elements of vector x). Since X a rank 4 matrix and cant be decomposed in such a way, I am looking for some approximate solution. Any suggestions on obtaining this vector x would be appreciated.

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Answer 1

John D'Errico 2015-11-18

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编辑：John D'Errico 2015-11-18

在 MATLAB Online 中打开

I'm sorry, but this is flat out impossible.

X = [0.4079 + 0.0000i   0.7532 + 0.0030i   0.9791 + 0.0272i   0.9335 - 0.0036i;
7532 - 0.0030i   1.2288 + 0.0000i   1.3074 + 0.0052i   0.9791 + 0.0272i;
9791 - 0.0272i   1.3074 - 0.0052i   1.2288 + 0.0000i   0.7532 + 0.0030i;
9335 + 0.0036i   0.9791 - 0.0272i   0.7532 - 0.0030i   0.4079 + 0.0000i];

Not even close to being possible. Lets see why.

rank(X)
ans =
     4
svd(X)
ans =
       3.7714
      0.62027
      0.11221
     0.010048

Only if the rank of X was 1, i.e., it had 1 non-zero singular values, and 3 essentially zero values, could you do this.

The point is, to represent the matrix as an outer product of two vectors, i.e., x'*y, where x and y are row vectors, the result would have rank 1. That is a fundamental of linear algebra.

As you can see, that is clearly not true. Simply wanting to do the impossible is not an option. Case closed, IF you want an exact solution.

If you wish to find the closest approximation as such a result (based on reading your comments) then it is possible. Use the SVD.

[U,S,V] = svd(X);
u = U(:,1)
u =
   -0.41048 - 4.0811e-18i
   -0.57576 +   0.001811i
   -0.57571 +  0.0081111i
   -0.41042 +  0.0070737i
s = S(1,1)
s =
     3.7714
v = V(:,1)
v =
   -0.41048 +          0i
   -0.57576 +   0.001811i
   -0.57571 +  0.0081111i
   -0.41042 +  0.0070737i

Since X is symmetric, we see that u and v will be the same. If the goal is to write Xhat as a product w*w', then wejust use

w = u*sqrt(s)
w =
   -0.79716 - 7.9255e-18i
    -1.1181 +   0.003517i
     -1.118 +   0.015752i
   -0.79705 +   0.013737i
w*w'
ans =
    0.63547 +          0i      0.89134 +  0.0028036i      0.89125 +   0.012557i      0.63538 +   0.010951i
    0.89134 -  0.0028036i       1.2502 +          0i       1.2502 +   0.013681i      0.89125 +   0.012557i
    0.89125 -   0.012557i       1.2502 -   0.013681i       1.2502 +          0i      0.89134 +  0.0028036i
    0.63538 -   0.010951i      0.89125 -   0.012557i      0.89134 -  0.0028036i      0.63547 +          0i

As you can see, this must be rank 1.

rank(w*w')
ans =
     1

The norm of the error of approximation is as small as possible.

norm(w*w' - X)
ans =
      0.62027

So this is the best way (least squares) to produce a rank 1 approximation to X. By way of comparison,

norm(X)
ans =
       3.7714

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Torsten 2015-11-18

Since the OP wants to use w*w^H as approximation, he can use w=u*sqrt(s).

Best wishes

Torsten.

John D'Errico 2015-11-18

编辑：John D'Errico 2015-11-18

Fixed

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Answer 2

Torsten 2015-11-18

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编辑：Walter Roberson 2015-11-18

https://en.wikipedia.org/wiki/Low-rank_approximation

In your case, X=U*Sigma*U^H.

Consequently, u1*sigma1*u1^H where u1 is the eigenvector corresponding to the largest eigenvalue sigma1, is the best rank-1 - approximation to X in the Frobenius norm.

Best wishes

Torsten.