For the probability space $(\Omega, \mathscr F, \mathbb P)$, the random variable $X$ with distribution $F_X$ has Skorokhod representation
$$X(\omega) := \sup \{x \in \mathbb R | F_X(x) < \omega\}$$
It can be shown that in the probability space $(\mathbb R, \mathscr B (\mathbb R), \mathcal L_X)$, the identity random variable
$$Y(\omega) = \omega$$
has distribution $F_Y = F_X$.
Let's take the exponential distribution:
In $((0,1), \mathscr B(0,1), Leb)$, it looks like:
$$X(\omega) = \frac{1}{\lambda}\ln(\frac{1}{1-\omega})$$
or
$$X(1-\omega) = \frac{1}{\lambda}\ln(\frac{1}{\omega})$$
Generally, in $(\Omega, \mathscr F, \mathbb P)$, it looks like:
$$X(\omega) := \sup \{x \in \mathbb R | F_X(x) < \omega\}$$
Specifically, in $(\mathbb R, \mathscr B (\mathbb R), 1-e^{-\lambda x})$, it looks like:
$$X(t) := t$$
The expectations for each are
$$\lambda^{-1} = \int_0^1 \frac{1}{\lambda}\ln(\frac{1}{1-\omega}) d\omega = \int_0^1 \frac{1}{\lambda}\ln(\frac{1}{\omega}) d\omega$$
$$\lambda^{-1} = \int_{\Omega} X d \mathbb P = \int_{\Omega} \sup \{x \in \mathbb R | F_X(x) < \omega\} d \mathbb P(\omega)$$
$$\lambda^{-1} = \int_{\mathbb R} t dF_X(t) = \int_{\mathbb R} t d(1-e^{-\lambda t}) = \int_{\mathbb R} t f_X(t) dt = \int_{\mathbb R} t \lambda e^{-\lambda t} dt$$
I'm giving you a quick answer because I'm lazy, but hope this helps you!
Treat the $m \times n$ matrix as a big vector of dimensions $mn$. So, treat, $X$ as a random vector $X: \Omega \to \mathbb{R}^{mn}$. Now, look the this wiki page:
https://en.wikipedia.org/wiki/Multivariate_random_variable
In short, $E(X)$ will be a matrix of same dimensions so that $E(X)_{ij}:=E(X_{ij}) \in \mathbb{R}$. Now, following the wiki page, the covaraince of your random variable will be a $mn \times mn$ dimensional matrix.
Best Answer
Let $(\Omega,\mathcal{A},P)$ be a probabilty space, $(\mathbb{R},\mathcal{B})$ a measurable space and $X\colon \Omega\to\mathbb{R}$ a random variable. The expected value of $X$ is defined as the Lebesgue integral \begin{align} \mathbb{E}_P(X):=\int_{\Omega}X\,\mathrm{d}P \end{align} and by the change of variables formula it holds \begin{align} \int_{\Omega}X\,\mathrm{d}P=\int_{\mathbb{R}}x\,\mathrm{d}P_{X} \end{align} where $P_X:=P\circ X^{-1}$ is the distribution (push forward, image measure, $\ldots$) of $X$ with respect to $P$. If $X$ has a probabilty density function $f=\frac{\mathrm{d}P_X}{\mathrm{d}\lambda}$, we can write \begin{align} \int_{\mathbb{R}}x\,\mathrm{d}P_{X}=\int_{\mathbb{R}}x\cdot f(x)\;\mathrm{d}\lambda(x) \end{align}