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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;

0.7532 - 0.0030i 1.2288 + 0.0000i 1.3074 + 0.0052i 0.9791 + 0.0272i;

0.9791 - 0.0272i 1.3074 - 0.0052i 1.2288 + 0.0000i 0.7532 + 0.0030i;

0.9335 + 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.

John D'Errico
on 18 Nov 2015

Edited: John D'Errico
on 18 Nov 2015

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;

0.7532 - 0.0030i 1.2288 + 0.0000i 1.3074 + 0.0052i 0.9791 + 0.0272i;

0.9791 - 0.0272i 1.3074 - 0.0052i 1.2288 + 0.0000i 0.7532 + 0.0030i;

0.9335 + 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

Torsten
on 18 Nov 2015

Edited: Walter Roberson
on 18 Nov 2015

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.

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