Does the correlation between the measurement noise influence the result of the LS?

Question

Consider a linear measurement process with some noise: $$y=Hx+v$$ with $$v \sim \mathcal{N}(0,\Sigma)$$ the covariance matrix $\Sigma$ is not a diagonal matrix. As we know, using LS, the $\hat{x}$ is $$\hat{x} = (H^T H)^{-1} H^T y$$ Does it change when $\Sigma$ is not a diagonal matrix?

Answer 1

Sorry, I eventually realize it is a simple question. Here is the answer of mine.

first, for a semi-definite matrix $\Sigma$, we can find $\Lambda$ so that $$\Sigma = \Lambda \Lambda^T$$

Then, we have $$\Lambda^{-1} v ∼ \mathcal{N}(0,I) $$

So, we can apply the conventional LS to modified measurement process: $$ \Lambda^{-1}y =\Lambda^{-1}Hx+\Lambda^{-1}v$$ then $$\hat{x} = ((\Lambda^{-1}H)^T \Lambda^{-1}H)^{-1} (\Lambda^{-1}H)^T \Lambda^{-1} y$$ $$\hat{x} = (H^T \Sigma ^{-1}H)^{-1} H^T \Sigma ^{-1} y$$