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"

  • $\begingroup$ 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 $(N+M)\times(N+M)$? $\endgroup$ Oct 6, 2015 at 17:10
  • $\begingroup$ @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. $\endgroup$
    – user34626
    Oct 6, 2015 at 18:12
  • $\begingroup$ after the edit it looks like $C$ has size $(2M-1)\times(2M-1)$; where does $N$ enter? $\endgroup$ Oct 6, 2015 at 18:29
  • $\begingroup$ @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. $\endgroup$
    – user34626
    Oct 6, 2015 at 18:39

1 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:

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

  • $\begingroup$ 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. $\endgroup$
    – user34626
    Oct 6, 2015 at 20:34
  • $\begingroup$ $Y=\frac{1}{2}\sigma ^2N\text{tr}\mathbf{AA}^H$ does not match with simulation. $\endgroup$
    – user34626
    Oct 6, 2015 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – user34626
    Oct 7, 2015 at 22:00
  • $\begingroup$ 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. $\endgroup$
    – user34626
    Oct 7, 2015 at 22:44

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