# Is there a way to simplify the following trace expression?

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?

• it would seem that your matrix $C$ has size $(2n+1)\times(2n+1)$ --- is that what you want? if it is, how does this relate to the size $(N+M)\times(N+M)$? Oct 6, 2015 at 17:10
• @CarloBeenakker Sorry this part was not clear, I updated the question. $[c_0,...,c_{M-1}]$ has length $M$ and the rest of rows and columns are filled with $N$ zeros in the matrix. Oct 6, 2015 at 18:12
• after the edit it looks like $C$ has size $(2M-1)\times(2M-1)$; where does $N$ enter? Oct 6, 2015 at 18:29
• @CarloBeenakker There are $N$ zeros between $c_0$ and $c_{M-1}$ in the first row. I guess the confusion comes by thinking that $M-1$ is the index of $c_{M-1}$ in the first row of the matrix. I agree it's confusing, but I cannot think of any other way to write it. The way I'm constructing the Toeplitz matrix is I'm taking an $M$ long vector $[c_1,...,c_{M-1}]$ and shifting it through every row of the matrix while filling the remaining $N$ spots in every row with zeros. Oct 6, 2015 at 18:39

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:

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

• Thank you for the answer. I'm trying to match the analytical solution you gave with simulation, currently they don't match. I'm trying to figure out why. Oct 6, 2015 at 20:34
• $Y=\frac{1}{2}\sigma ^2N\text{tr}\mathbf{AA}^H$ does not match with simulation. Oct 6, 2015 at 20:43
• I edited the question to add $M<N$, also made a small correction to the conditions for $N_k$ to reflect that. Answer matches perfectly with simulation. Thank you. Oct 7, 2015 at 22:00
• I don't think we need the "modulo" anymore: $\delta'_{ij}=1$ if $i=j\in\{0,1,2,\ldots M-1\}$ $\text{modulo}\,(N+M)$. I believe just $\delta'_{ij}=1$ if $i=j\in\{0,1,2,\ldots M-1\}$ will suffice. Oct 7, 2015 at 22:44