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
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?
