Is there a way to simplify the following trace expression?

Question

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"

Answer 1

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:

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