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Suppose the complex column vector $\mathbf{x}$ is linearly transformed by the complex matrix $\mathbf{T}$ into $\mathbf{y}$:

\begin{align} \mathbf{y} = \mathbf{T}{x} \end{align}

Assuming $\mathbf{T}$ is full rank, how could I estimate matrix $\mathbf{T}$, having a set of values of the column vectors $\mathbf{x}$ and $\mathbf{y}$?

EDIT: My goal is to identify a reduced linear operator that produce the same results as another operator with higher dimensions. I have a set of symbols that are hermitian symmetric around the $N/2$-th row, e.g., for $N=8$, $x[1] = x[7]^{*}$, $x[2] = x[6]^{*}$ and so on. Then, since this is an application from communications, assume that only the first half of the column vector carries useful information. Thus, I want to find a linear operator that involves only this half, with size $(N/2 +1) \times (N/2 +1)$, in order to avoid using too much memory and processing.

I have clear equations for the $N \times N$ original operator, but the operator with reduced dimensions is very hard to prove. Thus, I'm resorting to estimation to obtain conclusions about the possibility of coming up with such operator.

The estimation I'm currently implementing is the following (following Michael C. Grant's answer). 

SET_LENGTH = 1e3;

x =  fft(rand(8,SET_LENGTH));
N = 8;

% Linear Operator
A = rand(N,N); %must be real-valued
w = exp(-1j*2*pi/N); % twiddle factor
W = w.^(repmat(0:N-1,N,1).*repmat(0:N-1,N,1).'); % DFT matrix
T = W*A*W';

% Hermitian symmetric symbols
y = T*x;

% Positive frequencies
xp = x(1:(N/2+1),:);
yp = y(1:(N/2+1),:);

% Estimate linear operator with reduced dimensions
Tp = (yp*xp')/(xp*xp')

% Compare results
Tp*xp(:,2) % Estimated
yp(:,2) % Original

Does the fact that the estimated value and the original are significantly differet mean that my problem has no solution?

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igorauad
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  • This is very similar to this question: http://math.stackexchange.com/questions/391673 – Michael Grant May 20 '13 at 16:05
  • Thank you very much. It is very good to hear the answer from you, since I am trying to use CVX. I edited the main question with a routine. I would appreciate to know if there is a better way of finding this solution. – igorauad May 21 '13 at 00:08
  • There's actually no need to use CVX. The link I posted above uses simple linear systems. For instance, if there are exactly $N$ vectors, $T=YX^{-1}$, where $Y$ and $X$ contain the vectors collected as columns. – Michael Grant May 21 '13 at 00:40
  • Otherwise (if the are more than $N$ vectors in the set), should I use $W = YX^H(XX^H)^{-1}$? Once again, thanks for the support. – igorauad May 21 '13 at 02:23
  • Yes, indeed, that is right. Now go over to that link and give be an upvote, eh? ;-) – Michael Grant May 21 '13 at 02:29
  • Thank you very much. My reputation is so low that I'm not allowed to give an upvote. – igorauad May 21 '13 at 13:00
  • No worries then :-) – Michael Grant May 21 '13 at 13:08
  • @MichaelC.Grant, since I posted this question, I've been trying to formulate the closed form of the operator. However, because I haven't had success yet, I decided to reevaluate what the estimation says to me. The error between estimation and original is significant. Am doing wrong or, because the vectors and the linear operator are complex, the implementation is different? (The original post was edited to present the .m code I'm using ). Thank you very much in advance) – igorauad May 27 '13 at 15:43
  • I'm not sure what to recommend here. Try it with some manufactured problems that you know are correct to make sure that your math is right. You might have your matrix conjugated or something like that, that would throw you off. – Michael Grant May 27 '13 at 20:55
  • Actually, that was a good recommendation. I tried it with full $(N \times N)$ matrices, which I knew were right, and it worked perfectly. The question now is, can I conclude the problem does not have solution, by looking at the error on estimation? Thank you very much. – igorauad May 28 '13 at 16:08

0 Answers0