How to make colored noise white? I am studying linear minimum mean squared error (MMSE) receivers. The technique used in it suggests to make colored noise white by multiplying it with the invertible covariance matrix $K_z^{-\frac{1}{2}}$, with $K$  given by 

$$K_{zk} = N_0 I_nr +\sum_{i≠k}^{n_t}P_ih_ih_i^*$$

where $h_i$ and $h_i^*$ are the channel gain and channel gain conjugate transpose, respectively, and $n_t$ is the number of transmit antennas. The colored noise is because of interference from other users (I guess so!).

I am confused as how the multiplication of $y = hx + z$ with $K_z^{-\frac{1}{2}}$ can transform colored noise to white though I can guess that may be multiplying with inverse cancel out the colored part and the noise is white now. Am I correct?
Given the linear model $$y = hx + z$$ where $z \sim \mathcal{N}(0,K_z)$. We say that noise is colored when $K_z \neq \Sigma$, a diagonal matrix. To uncolor the noise, you can pre-multiply $y$ by $K_z^{-\frac{1}{2}}$, so 
$$K_z^{-\frac{1}{2}}y = K_z^{-\frac{1}{2}}hx +K_z^{-\frac{1}{2}}z = K_z^{-\frac{1}{2}}hx +\tilde{z}$$
Notice that the noise is now $\tilde{z} = K_z^{-\frac{1}{2}}z$, which is uncolored since $$E(\tilde{z} \tilde{z} ^T) =  K_z^{-\frac{1}{2}}E({z} {z} ^T)K_z^{-\frac{1}{2}} = K_z^{-\frac{1}{2}}K_zK_z^{-\frac{1}{2}} =  K_z^{-\frac{1}{2}}K_z^{+\frac{1}{2}}K_z^{+\frac{1}{2}}K_z^{-\frac{1}{2}} = I$$
So $\tilde{z} \sim \mathcal{N}(0,I)$