Proving that Covariance of residuals and errors is zero This is exercise 9.8 from John Fox's "Applied Regression Analysis & Generalized Linear Models". The question text is: 
The statistic: 
$$
t = \frac{B_j - \beta_j}{ S_E \sqrt{v_{jj}} }
$$
to have a t-distribution, the estimators $B_j$ and $S_E$ must be independent. [Here, $v_{jj}$ is the jth diagonal entry of $(X^{\top}X)^{-1}$.] The coefficient $B_j$ is the $j$th element of $\vec{b}$, and $S_E = \sqrt{\vec{e}^\top\vec{e}/(n-k-1)}$ is a function of the residuals $\vec{e}$. Because both $\vec{b}$ and $\vec{e}$ are normally distributed, it suffices to prove that their covariance is $\vec{0}$. Demonstrate that this is the case. [Hint: Use $\operatorname{Cov}(\vec{e},\vec{b}) = \mathbb{E}[\vec{e}(\vec{b} - \vec{\beta})^\top]$ and begin by showing that $\vec{b} - \vec{\beta} = (X^\top X)^{-1} X^\top\epsilon$.]
So that is the question. I have shown that $\vec{b} - \vec{\beta} = (X^{\top}X)^{-1}X^{\top}\epsilon$, and after that is where I am stuck. Here is what I have done so far: 
Note: $\vec{\epsilon}$ is the statistical error in linear regression model $Y = X\beta + \epsilon$, and $\vec{e}$ is the vector of residuals. $\vec{b}$ is the ordinary least squares estimator for $\vec{\beta}$, the vector of regressor coefficients.  
$$
\begin{aligned}
\operatorname{Cov}(\vec{e},\vec{b}) &= \mathbb{E}[\vec{e}(\vec{b} - \vec{\beta})^{\top}]\\
    &= \mathbb{E}[\vec{e} ((X^\top X)^{-1}X^\top\vec{\epsilon})^\top]  \\
    &= \mathbb{E}[\vec{e} \vec{\epsilon}^\top X(X^\top X)^{-1}]  \\
    &= \mathbb{E}[\vec{e} \vec{\epsilon}^\top]X(X^\top X)^{-1}
\end{aligned}
$$
But I am not sure how to proceed after this. I'm not sure what to do with $\mathbb{E}[\vec{e} \vec{\epsilon}^\top]$. I was wondering if those two variables are independent, and if so I could split that expectation into the product of the expectations, and then use $\mathbb{E}[\vec{\epsilon}] = 0$, since one of the assumptions of regression (at least in this context) is that the error terms are i.i.d. normal with mean zero. But I'm not sure I can justify that the two terms are independent or if that's even true at all. 
 A: I think I got the answer. I'd appreciate if anyone could confirm this is correct, and if not help me out!
$$
\begin{aligned}
 Cov(\vec{e},\vec{b}) &= \mathbb{E}[\vec{e}(\vec{b} - \vec{\beta})^{\top}]\\
  &= \mathbb{E}[\vec{e} ((X^{\top}X)^{-1}X^{\top}\vec{\epsilon})^{\top}] \\ 
  &= \mathbb{E}[\vec{e} \vec{\epsilon}^{\top}X (X^{\top}X)^{-1}] \\ 
  &= \mathbb{E}[(\vec{y}-\vec{\hat{y}}) \vec{\epsilon}^{\top}X (X^{\top}X)^{-1}] \\ 
  &= \mathbb{E}[(X\vec{\beta} + \vec{\epsilon} -X\vec{b}) \vec{\epsilon}^{\top}X (X^{\top}X)^{-1}] \\ 
  &= \mathbb{E}[X\vec{\beta}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} + \vec{\epsilon}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} -  X\vec{b}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} ] \\ 
  &= \mathbb{E}[X\vec{\beta}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} + \vec{\epsilon}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} -  X((X^{\top}X)^{-1}X^{\top}\vec{y})\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} ] \\ 
  &= \mathbb{E}[X\vec{\beta}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} + \vec{\epsilon}\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} -  X(X^{\top}X)^{-1}X^{\top}(X\beta + \epsilon)\vec{\epsilon}^{\top}X (X^{\top}X)^{-1} ] \\ 
  &= \mathbb{E}[ X\vec{\beta}\vec{\epsilon}^{\top} + \vec{\epsilon}\vec{\epsilon}^{\top} -  X(X^{\top}X)^{-1}X^{\top}(X\beta + \epsilon)\vec{\epsilon}^{\top} ]X (X^{\top}X)^{-1} \\ 
  &= \mathbb{E}[ X\vec{\beta}\vec{\epsilon}^{\top} + \vec{\epsilon}\vec{\epsilon}^{\top} -  X(X^{\top}X)^{-1}X^{\top}X\beta\vec{\epsilon}^{\top} - X(X^{\top}X)^{-1}X^{\top}\epsilon\vec{\epsilon}^{\top} ]X (X^{\top}X)^{-1} \\  
  &= \mathbb{E}[ X\vec{\beta}\vec{\epsilon}^{\top} + \vec{\epsilon}\vec{\epsilon}^{\top} -  X\beta\vec{\epsilon}^{\top} - X(X^{\top}X)^{-1}X^{\top}\epsilon\vec{\epsilon}^{\top} ]X (X^{\top}X)^{-1} \\ 
  &= \mathbb{E}[ \vec{\epsilon}\vec{\epsilon}^{\top} - X(X^{\top}X)^{-1}X^{\top}\epsilon\vec{\epsilon}^{\top} ]X (X^{\top}X)^{-1} \\ 
  &= \mathbb{E}[ (I - X(X^{\top}X)^{-1}X^{\top})\epsilon\vec{\epsilon}^{\top} ]X (X^{\top}X)^{-1} \\ 
  &= (I - X(X^{\top}X)^{-1}X^{\top})\mathbb{E}[ \epsilon\vec{\epsilon}^{\top} ]X (X^{\top}X)^{-1} \\ 
  &= (I - X(X^{\top}X)^{-1}X^{\top})\sigma^{2}IX (X^{\top}X)^{-1} \\ 
  &= \sigma^{2}(I - X(X^{\top}X)^{-1}X^{\top})X (X^{\top}X)^{-1} \\ 
  &= \sigma^{2}(X (X^{\top}X)^{-1} - X(X^{\top}X)^{-1}X^{\top}X (X^{\top}X)^{-1}) \\ 
  &= \sigma^{2}(X (X^{\top}X)^{-1} - X(X^{\top}X)^{-1}) \\ 
  &= \sigma^{2}(\mathbf{0}) \\ 
  &= \mathbf{0}
\end{aligned}
$$
