Say that the discrete Fourier transform (DFT) is used in OFDMA. There are a number of degenerate (singular, non invertible) sub matrices of some DFT matrices. Does this result in any problems?

One example from FFT is where some basis vectors have some elements in common like in this size 8 DFT/FFT. Let us express the DFT as a matrix multiplication X = M*x

Size 8 DFT.

Say we have an input vector with a non zero sum of x[0] and x[4] and other elements zero. Output X[0], X[2] and X[4] will have the same value regardless of how the sum is distributed between x[0] and x[4]. This does not look good. It is the same as saying that the submatrices

$$ \begin{bmatrix}M_{00}&M_{04}\\M_{20}&M_{24}\end{bmatrix} = \begin{bmatrix}M_{20}&M_{24}\\M_{40}&M_{44}\end{bmatrix} = \begin{bmatrix}M_{00}&M_{04}\\M_{40}&M_{44}\end{bmatrix} = \begin{bmatrix}1&1\\1&1\end{bmatrix} $$ are degenerate, having determinant zero. Similar sub degeneracies exist for larger $2^n$ sample sizes. The entire resulting vector X of-course contains all information in the original vector x but the meaning with some applications of OFDMA is to use different parts of the frequency domain for different transceivers.

X is then transmitted, and some noise is added, and an inverse FFT is performed. It has in this example the matrix formulation x = $M^{-1}$ X

$$ M^{-1} = \frac{1}{8}\begin{bmatrix} 1&1&1&1&1&1&1&1\\ 1&\frac{1}{\sqrt2}(1+i)&i&\frac{1}{\sqrt2}(-1+i)&-1&\frac{1}{\sqrt2}(-1-i)&-i&\frac{1}{\sqrt2}(1-i)\\ 1&i&-1&-i&1&i&-1&-i\\ 1&\frac{1}{\sqrt2}(-1+i)&-i&\frac{1}{\sqrt2}(1+i)&-1&\frac{1}{\sqrt2}(1-i)&i&\frac{1}{\sqrt2}(-1-i)\\ 1&-1&1&-1&1&-1&1&-1\\ 1&\frac{1}{\sqrt2}(-1-i)&i&\frac{1}{\sqrt2}(1-i)&-1&\frac{1}{\sqrt2}(1+i)&-i&\frac{1}{\sqrt2}(-1+i)\\ 1&-i&-1&i&1&-i&-1&i\\ 1&\frac{1}{\sqrt2}(1-i)&-i&\frac{1}{\sqrt2}(-1-i)&-1&\frac{1}{\sqrt2}(-1+i)&i&\frac{1}{\sqrt2}(1+i)\\ \end{bmatrix} $$

Some degenerate sub matrices, there are even more, are

$$ \begin{bmatrix}{M^{-1}}_{00}&{M^{-1}}_{04}\\{M^{-1}}_{20}&{M^{-1}}_{24}\end{bmatrix} = \begin{bmatrix}{M^{-1}}_{20}&{M^{-1}}_{24}\\{M^{-1}}_{40}&{M^{-1}}_{44}\end{bmatrix} = \begin{bmatrix}{M^{-1}}_{00}&{M^{-1}}_{04}\\{{M^{-1}}^{-1}}_{40}&{M^{-1}}_{44}\end{bmatrix} = \begin{bmatrix}1/8&1/8\\1/8&1/8\end{bmatrix} $$

so for noise where X[0]+X[2]+X[4] = constant, x[0] and x[4] stays the same. Doesn't this change the signal to noise ratio or spectral efficiency?

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    $\begingroup$ @MarcusMüller That is the question I am asking. Maybe there is no problem, but there is a difference. If I were to hint for an answer wouldn't that be a risk to bias an answer? Someone could write an answer just stating "No" and people could vote for it. But did you notice the tag spectral-efficiency? We have three frequencies not showing any information about x[0]-x[4]. Is this never an issue? What if we have a desired signal and a noise of x[0]+x[4]? $\endgroup$ Commented Feb 6, 2021 at 1:03
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    $\begingroup$ @MarcusMüller Point taken. Better now? Or should I specify some other domain than OFDM? $\endgroup$ Commented Feb 6, 2021 at 1:35
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    $\begingroup$ @DavidJonsson I do not agree with " We have three frequencies not showing any information about x[0]-x[4]". Without a more specific example that can be considered as an "issue", my answer would be that as the DFT is a reversible transform, any usage of the submatrices of the DFT-matrix form representation will result in ambiguity. It is what it is and someone can call it an issue, why not $\endgroup$
    – AlexTP
    Commented Feb 6, 2021 at 10:59
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    $\begingroup$ @AlexTP Not all DFTs have this property, but all FFTs of size bigger than 4. $\endgroup$ Commented Feb 8, 2021 at 21:56
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    $\begingroup$ @MarcusMüller You have apparently missed the point of the question. I mention singular submatrices. Just because a DFT is invertible doesn't mean that any submatrix from it is. Only prime sized DFTs have all sub matrices invertible. $\endgroup$ Commented Sep 24, 2021 at 14:21


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