I'd like to simplify the following expression:

$$\text{tr}\{\mathbf{A}^HE(\mathbf{C}^H \begin{bmatrix} \mathbf{0}_{M\times M} & \mathbf{0}_{M\times N} \\ \mathbf{0}_{N\times M} & \mathbf{I}_{N\times N} \end{bmatrix} \mathbf{C})\mathbf{A}\}$$, where the matrix $\mathbf{C}$ is Toeplitz and is constructed by shifting the vector $[c_0,\cdots, c_{M-1}]$ through the rows while filling the rest of elements in every row with $N$ zeros, $M<N$.

$$\mathbf{C} = \begin{bmatrix} c_0 & 0 & \cdots & 0 & c_{M-1} & \cdots & c_1\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & c_{M-1}\\ c_{M-1} & \cdots & \cdots & c_0 & 0 & \cdots & 0\\ 0 & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots\\ 0 & \ddots & 0 & c_{M-1} & \cdots & \cdots & c_0 \end{bmatrix}~,$$ $\mathbf{A} \in \mathbb{C}^{(N+M)\times N}$, $\mathbf{C} \in \mathbb{C}^{(N+M)\times (N+M)}$ and $E(c_ic_i^*)=\frac{\sigma^2}{2}$, $E(c_ic_j^*)=0$. $\text{tr}(\cdot)$ is the trace, $E(\cdot)$ is the expectation, $(\cdot)^H$ is the hermitian and $(\cdot)^*$ is the conjugate. $\mathbf{I}$ and $\mathbf{0}$ are the identity and zero matrices respectively.

In particular, I want to get rid of the expectation operator, any suggestions?

Cross posted at "https://math.stackexchange.com/questions/1466196/is-there-a-way-to-simplify-the-following-trace-expression"