In the simplest scenario of MIMO spatial multiplexing: $$\mathbf{y} = H\mathbf{s} + \mathbf{n}$$ where: $\mathbf{s}=[s_0,s_1,...s_{M-1}] \\\mathbf{y}=[y_0,y_1,...y_{N-1}]$
$\mathbf{n}=[n_0,n_1,...n_{N-1}]$ is zero mean complex normal independent Random vector
$M$ transmit antennas, $N$ receive antennas where $N\geq M$
The maximum likelihood solution is: $$\tilde{\mathbf{s}} = \arg\min_{\mathbf{s} \in QPSK^M} ||\mathbf{y}-H\mathbf{s}||^2$$
Now, my textbook says that due to the fact that in this scenario $H^{*}H$ is not a diagonal matrix, the least squares solution followed by proccesing is not optimal, but I can't see why this is true.
To my understanding, denoting the LS solution as $\hat{s}=(H^{*}H)^{-1}H^{*}\mathbf{y}$
I can write:
$$ ||\mathbf{y}-H\mathbf{s}||^2 = ||\mathbf{y}-H\hat{s}+H\hat{s}-H\mathbf{s}||^2 = \\||y-H\hat{s}||^2 + ||H(\hat{s}-s)||^2 + 2Re[(y-H\hat{s})^{*}H(\hat{s}-s)]$$
Now, by substituing the LS solution the $Re$ part vanishes. And the problem reduces to $$\arg\min_{s\in QPSK^{M}}|\hat{s}-s|^2$$ So I don't get why the ML is not optimal is this case. Can anyone please clarify? Thanks