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>"



Is there a way to simplify the following trace expression?
After a cyclic permutation of the trace, the expression you need is


$$Y=\text{tr}\left\{\mathbf{A}^HE(\mathbf{C}^H \begin{bmatrix} \mathbf{0}\_{M\times M} & \mathbf{0}\_{M\times N} \\ \mathbf{0}\_{N\times M} & \mathbf{I}\_{N\times N} \end{bmatrix} \mathbf{C})\mathbf{A}\right\}=\text{tr}\left\{E(\mathbf{C}\mathbf{A}\mathbf{A}^H \mathbf{C}^H) \begin{bmatrix} \mathbf{0}\_{M\times M} & \mathbf{0}\_{M\times N} \\ \mathbf{0}\_{N\times M} & \mathbf{I}\_{N\times N} \end{bmatrix} \right\}$$


Let me abbreviate $\mathbf{H}=\mathbf{AA}^H$, $\mathbf{X}=\mathbf{C}\mathbf{A}\mathbf{A}^H \mathbf{C}^H$, and let me denote by $\mathbf{P}$ the projection matrix in square brackets, so $Y={\rm tr}\{E(\mathbf{X})\mathbf{P}\}$. The expectation value can be evaluated using the Toeplitz property $C\_{ij}=c\_{i-j}$. 


From your construction I gather that
$$E(c\_i c^\ast\_j)=\tfrac{1}{2}\sigma^2\delta'\_{ij}$$
where $\delta'\_{ij}=1$ if $i=j\in\{0,1,2,\ldots M-1\}$ $\text{modulo}\,(N+M)$, while $\delta'\_{ij}=0$ otherwise.


Carrying out the average,
$$E(X\_{ij})=\sum\_{kl}E(C\_{ik}H\_{kl}C^\ast\_{jl})=\sum\_{kl}E(c\_{i-k}H\_{kl}c^\ast\_{j-l})=\tfrac{1}{2}\sigma^2\sum\_{kl}H\_{kl}\delta'\_{i-k,j-l}$$


The projector $\mathbf{P}$ identifies $j=i=M+1,M+2,\ldots M+N$, so we arrive at


$$Y={\rm tr}\{E\mathbf{(X)P}\}=\tfrac{1}{2}\sigma^2 \sum\_{i=M+1}^{N+M}\sum\_{k,l=1}^{N+M}H\_{kl}\delta'\_{i-k,i-l}=\tfrac{1}{2}\sigma^2 \sum\_{k=1}^{N+M}H\_{kk}N\_k$$


with the definition $N\_k=\sum\_{i=M+1}^{N+M}\delta'\_{i-k,i-k}$. This number is a bit tedious to evaluate [\*], but I presume you can easily take it from here.


[\*] If I have not made a mistake, I find:


$$N\_k=
\begin{cases}
k-1& \text{if}\quad 1\leq k\leq M\\
M& \text{if}\quad M+1\leq k\leq N+1\\
N+M-k+1& \text{if}\quad N+2\leq k\leq M+N\\
0&\text{otherwise}
\end{cases}$$


