Simplification of an expectation While attempting to simplify a combination of expectations, I'm stuck at a particular term whose simplification I'm unable to deduce.
The term to be simplified is: $\mathbb{E}[X^{T}F^{T}FX]$
where $F$ is a given matrix of $\dim(p,n)$ while $X$ is a $\dim(n,p)$ matrix where each of its rows is a vector generated from the distribution $\mathcal{N}(0,K)$ and $K$ is the covariance matrix.
In other words, the rows ${R_1, R_2 \ldots R_n}$ of $X$ are random vector that follows the distribution $\mathcal{N}(0,K)$.
Any help / hint in order to simplify the expectation in terms of $F$ and $K$ would be very useful for my work.
 A: Your stochastic matrix $X$ have each row a vector with multivariate normal distribution $\mathcal{N}(0, K)$. You did not specify if the rows $R_j^T$ are independent, but I will assume that. 
So let $X =\begin{pmatrix} R_1^T \\ R_2^T \\ \vdots \\ R_n^T\end{pmatrix}$ and $F^T=\begin{pmatrix} f_1^T \\f_2^T \\ \vdots \\ f_n^T \end{pmatrix}$.Then we can write
$$
 FX = \begin{pmatrix} f_1 & f_2 & \dots & f_n \end{pmatrix}\begin{pmatrix} R_1^T \\ R_2^T \\ \vdots \\ R_n^T\end{pmatrix}=\sum_{j=1}^n f_j R_j^T
$$ and using this
$$
X^T F^T F X= \sum_j R_j f_j^T \cdot \sum_k f_k R_k^T =\sum_j\sum_k R_j f_j^T f_k R_k^T
$$
note that $f_j^T f_k$ is a number, and by the assumed independence (and expectation zero of $R_j$) the terms for $j \not= k$ are zero.
Then we can conclude that
$$ \DeclareMathOperator{\E}{\mathbb{E}}\DeclareMathOperator{\trace}{\mathrm{trace}}
\begin{align}
\E X^T F^T F X &= \sum_j \E R_j f_j^T f_j R_j^T =\sum_j f_j^T f_j \E R_j R_j^T \\
&= K \sum_j f_j^T f_j = K \trace F^T F
\end{align}
$$
There is a superficial similarity to the question Confused about finding expectation of vector multiplied by matrix, but the trace trick used there cannot be used in this case. 
