0
$\begingroup$

In the paper I am referring (and here from citeseer), complex vectors $\mathbf{z}$ and matrix $\mathbf{M}$ were defined as follows

\begin{align} {{\bf z}} &= \left[z_{1},z_{2},\ldots,z_{N}\right]^{T} \in {\mathbb{C}}^{N} \\ {\tilde{\bf z}} &= \left[z_{1},z^{*}_{1},\ldots, z_{N},z^{*}_{N}\right]^{T} \in {\mathbb{C}}^{2N} \\ {\hat{\bf z}} &= \left[z^{R}_{1},z^{I}_{1},\ldots, z^{R}_{N},z^{I}_{N}\right]^{T} \in {\mathbb{R}}^{2N}\\ {\bf M} &\in {\mathbb{C}}^{N \times N}\cr \tilde{{\bf M}} & \in {\mathbb{C}}^{2N \times 2N} \cr {\hat{\bf M}} & \in {\mathbb{R}}^{2N \times 2N} \end{align}

However, I can't seem to figure out how $\tilde{\mathbf{M}}$ would look like.

I have\begin{equation} \mathbf{M} = \begin{bmatrix} z_{11} & z_{12} & \dots & z_{1N}\\ z_{21} & z_{22} & \dots & z_{2N}\\ \vdots & \vdots & \ddots & \vdots\\ z_{N1} & z_{N2} & \dots & z_{NN} \end{bmatrix} \end{equation}

and the paper later clarifies

\begin{equation} \hat{\mathbf{M}} = \begin{bmatrix} z_{11}^R & -z_{11}^I & \dots & z_{1N}^R & - z_{1N}^I\\ z_{11}^I & z_{11}^R & \dots & z_{1N}^I & z_{1N}^R\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ z_{N1}^R & -z_{N1}^I & \dots & z_{NN}^R & - z_{NN}^I\\ z_{N1}^I & z_{N1}^R & \dots & z_{NN}^I & z_{NN}^R\\ \end{bmatrix} \end{equation}

but not for $\tilde{\mathbf{M}}$.

Thank you very much!

$\endgroup$

1 Answer 1

2
+50
$\begingroup$

The matrices $\hat{\mathbf{M}}$ and $\tilde{\mathbf{M}}$ are constructed in such a way that the relation $\mathbf{M}\mathbf{x}=\mathbf{y}$ implies $\hat{\mathbf{M}}\hat{\mathbf{x}}=\hat{\mathbf{y}}$ and $\tilde{\mathbf{M}}\tilde{\mathbf{x}}=\tilde{\mathbf{y}}$.

Consequently, for constructing the matrix $\tilde{\mathbf{M}}$, each element $m_{kl}$ of $\mathbf{M}$ must be replaced by a $2\times 2$ sub-matrix

$$\begin{bmatrix}m_{kl}&0\\0&m_{kl}^*\end{bmatrix}$$

In a similar way, the matrix $\hat{\mathbf{M}}$ is formed by replacing each element $m_{kl}$ of $\mathbf{M}$ by the $2\times 2$ sub-matrix

$$\begin{bmatrix}m_{kl}^R&-m_{kl}^I\\m_{kl}^I&m_{kl}^R\end{bmatrix}$$

where the superscripts $^R$ and $^I$ denote real and imaginary parts, respectively.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.