Question about sample covariance matrix

[sanctifier]

Suppose vectors X₁, X₂, ... , X_n whose components are random variables are mutually independent(I mean X_i's are vectors of components with constants which are possible values of random variables labeled by the component indice, and all these labeled random variables are organized as a vector X, hence X_i's just are samples of such a X), and the sample mean is $\hat{M}$ = $\frac{1}{N}$$\sum$ _{$\stackrel{N}{i = 1}$} X_i, and the true mean of all X_i's is M. Then to estimate the covariance matrix of X_i, we employ the following formula:
$\hat{Ʃ}$ = $\frac{1}{N}$$\sum$ _{$\stackrel{N}{i = 1}$} {(X_i - $\hat{M}$)(X_i - $\hat{M}$)^T}
$\$$\$$\:$= $\frac{1}{N}$$\sum$ _{$\stackrel{N}{i = 1}$} {((X_i - M) - ($\hat{M}$ - M))((X_i - M) - ($\hat{M}$ - M))^T}
$\$$\$$\:$= $\frac{1}{N}$$\sum$ _{$\stackrel{N}{i = 1}$} (X_i - M)(X_i - M)^T - ($\hat{M}$ - M)($\hat{M}$ - M)^T
My question is how does the equal sign hold in the last step?
I did some work about this question, first I note that the transpose is a llinear transformation, i.e., for two vectors V and U, (V + U)^T = V^T + U^T, then I realize that the following equation is legal.
(V - U)(V - U)^T = V$\!$V^T - VU^T - UV^T + UU^T
Let V = (X_i - M) and U = ($\hat{M}$ - M), the terms missing in the last step of $\hat{Ʃ}$ are -VU^T and -UV^T, OK, I know the entries of E[VU^T] actually are covariances of X_i and $\hat{M}$, and I assume they are all zero, consequently the terms -VU^T and -UV^T do miss because of taking the expectation on $\hat{Ʃ}$, but in the last step, they vanished before taking the expectation! Why?
Finally, I also notice that the sign of UU^T = ($\hat{M}$ - M)($\hat{M}$ - M)^T has been changed from + to -, how does this happen?

 

[sanctifier]

Ok, if 1/N can be envisaged as a approximate probability of each entry of VU^T, this can explain the vanishing of -VU^T and -UV^T without taking a expectation, but how to explain the sign change of UU^T occurred in the last step?