Why does variance-covariance matrix of $\hat{\beta}$ have transpose inside? In estimating vector $\hat{\beta}$ of linear regression, why does variance-covariance matrix of $\hat{\beta}$:
$$V(\hat{\beta})=E[(\hat{\beta} - E(\hat{\beta}))(\hat{\beta} - E(\hat{\beta}))^T]$$
have as the second term inside $E()$ the transpose?
Since I've found covariance to be expressed like (with no transpose):

 A: You quote the definition of covariance of two variables, while the form with transpose is the definition of covariance matrix. In first case we are talking about two random variables $X$ and $Y$,
$$ \operatorname{cov}(X,Y) = \operatorname{E}{\big[(X - \operatorname{E}[X])(Y - \operatorname{E}[Y])\big]} $$
while in the second case $\mathbf{X}$ is a vector of random variables $X_1,\dots,X_n$
$$ \mathbf{X} = \begin{bmatrix}X_1 \\ \vdots \\ X_n \end{bmatrix} $$
so we are talking about covariance between multiple variables in form of covariance matrix
$$ \Sigma=\mathrm{E}
\left[
 \left(
 \mathbf{X} - \mathrm{E}[\mathbf{X}]
 \right)
 \left(
 \mathbf{X} - \mathrm{E}[\mathbf{X}]
 \right)^{\rm T}
\right] $$
and the transpose appears in here because you are multiplying two vectors.
A: A covariance matrix is matrix-valued. Covariance of two random variables is an element in that matrix, i.e. a scalar. Hence the covariance formula that you list yields a scalar, while computing $xx^T$ for vector-valued $x$ yields a matrix.
A: Small example: Consider a 2x1
$$v = \begin{bmatrix}
7\\ 
6
\end{bmatrix}$$
Note that
$$v'v = \begin{bmatrix}
7&6
\end{bmatrix}\begin{bmatrix}
7\\ 
6
\end{bmatrix}$$
is a scalar (or 1x1 matrix?)
But
$$vv' = \begin{bmatrix}
7\\ 
6
\end{bmatrix} \begin{bmatrix}
7&6
\end{bmatrix}$$
is a matrix
