Covariance of $Z'Vb$ given that the rows of V are i.i.d. Suppose that we have the following entities
$$
\underbrace{Z}_{n\times k},\quad\underbrace{V}_{n\times L},\quad \underbrace{b}_{L\times 1}.
$$
$Z$ and $b$ are nonstochastic whereas we assume that the rows of $V$ are i.i.d. with each having mean $0$ and covariance matrix $\Omega$ (dimension: $L\times L$). I'm trying to justify that the following $k\times 1$ vector
$$
S=Z'Vb
$$
has covariance matrix 
$$
Z'Zb'\Omega b
$$
which has dimension $k\times k$ because the 5 quantities above have dimensions
$$
Z': k\times n;\quad Z: n\times k;\quad b':1\times L;\quad \Omega:L\times L;\quad b:L\times 1.
$$
It's clear that $S$ has zero mean so the covariance matrix is just $E(SS')$. I tried expanding $SS'$ but I couldn't simplify things to produce the expression above. Can someone lend a hand please?
Edit: The above result is claimed in passing on page 1030 in Moreira (2003). Full article is accessible here.
 A: I'd bet that there is a simpler way, but I couldn't simplify it.
Some slight changes in notation; and to aid readability, I write in lowercase bold the (column) vectors, in uppercase the matrices.
Let $$\underbrace{V}_{n\times \ell}=\pmatrix{{\bf v}_1^t\\
{\bf v}_2^t\\
\cdots\\
{\bf v}_n^t\\
}= ({\bf u}_1 \, {\bf u}_2 \cdots {\bf u}_\ell)$$
where ${\bf v_i}$ (transposed rows of $V$) are random iid vectors of length $\ell$, with $E({\bf v_i})=0$ , $E({\bf v_i} {\bf v_i}^t)= \Omega$, $E({\bf v_i} {\bf v_j}^t)= 0$ ($i\ne j$).
Regarding the columns ${\bf u_i}$, they also have zero mean, but they are not iid. Let the covariance be $T^{(i,j)}=E({\bf u_i}{\bf u_j}^t)$ ; then it can be checked that $$T^{(i,j)}= \Omega_{i,j} I_n \tag{1}$$
(To clarify: in the above $I_n$ is the $n$ identity matrix; $T^{(i,j)}$ is -for each $(i,j)$ pair- a matrix; $\Omega_{i,j}$ is a scalar, an element of the matrix $\Omega$)
Let $\underbrace{\bf a}_{(
n\times 1)}=\underbrace{V}_{n\times \ell} \, \, \underbrace{{\bf b}}_{\ell \times 1}= \sum_{i=1}^\ell {\bf u}_i \, b_i$
Then $E({\bf a})={\bf 0}$ and $$E({\bf a} \, {\bf a}^t)= \sum_{i,j} T^{(i,j)} b_i b_j= \left(\sum_{i,j} \Omega_{i,j}  b_i b_j \right) I_n = {\bf b}^t \, \Omega \,{\bf b} I_n \tag{2}$$ 
Now, let $\underbrace{{\bf s}}_{k \times 1}= Q V  {\bf b} = \underbrace{Q}_{k \times n} \,\, \underbrace{{\bf a}}_{n  \times 1} $. 
Then $$E({\bf s}{\bf s}^t)=Q E({\bf a}{\bf a}^t) Q^t= Q Q^t {\bf b}^t \Omega \,{\bf b}\tag{3}$$
