How can a linear operator on DFT vector produce the same vector using only half of the DFT vector?
I've suggested in the comments above (and I think Matt L is trying to say the same thing also) that you should be able to just discard the negative frequency elements entirely for your analysis. They don't seem to be relevant to what you're doing (which makes sense, since they don't carry any information, as the time-domain signal is real). Look at your equation again:

$$
\mathbf{y} = \mathbf{Tx}
$$

(I changed the casing so that lower-case symbols are vectors and upper-case are matrices, as is the typical convention)

Assume for the time being that $\mathbf{x}$ and $\mathbf{y}$ are length-$N$ and thus $\mathbf{T}$ is $N$-by-$N$. This would correspond to the general case, where the time-domain signals corresponding to $\mathbf{x}$ and $\mathbf{y}$ need not be real. If $\mathbf{y}$ is a column vector, then you can think of the matrix operation instead as:

$$
\mathbf{y} = \sum_{k=1}^N x_k\mathbf{t_k}
$$

where $x_k$ is the $k$-th element of the vector $\mathbf{x}$ and $\mathbf{t_k}$ is the $k$-th column of the matrix $\mathbf{T}$. My question at this point would be, instead of using the full $N$-length vector and $N$-by-$N$ matrix, why not just use the $N/2$ rows that contain information? Then, the calculation would turn into:

$$
\mathbf{y^+} = \sum_{k=1}^{N/2} x_k\mathbf{t_k}
$$

(neglecting the details of the boundary cases for the zeroth and Nyquist bins, which depend upon whether $N$ is even or odd)

The resulting vector $\mathbf{y^+}$ contains the positive frequencies for the desired result. You can then just symmetrically extend it to get a full length-$N$ vector that has the conjugate symmetry that you seek. This should make intuitive sense in the following ways:


If there is no information in the negative frequencies in the input vector $\mathbf{x}$ (which there isn't, since the entire vector is defined by its positive frequency components), then you shouldn't need to use them in your calculations.
If there is no information in the negative frequencies in the output vector $\mathbf{y}$ (which there isn't, since the entire vector is defined by its positive frequency components), then you shouldn't need to waste time calculating what their values would be.


This should free you from worrying about any bizarre constraints on the properties of the matrix $\mathbf{T}$. I think you're trying to think of the action of the precoder as too coupled with the OFDM modulation. Instead, think of it just as a preprocessing operation that happens on the encoded symbols before you assign them to frequency bins in the OFDM modulator's IDFT.