I have the vector $y = Dx$ where $D$ is a complex matrix with dimension $N \times N$, and $x$ is a complex vector of dimension $N \times 1$. If the vector $y_2 = [y'_N, y'_{N-1}, y'_{N-2},.... , y'_{1}]$, where $ y'_{N}$ is the conjugate of the element $ y_N$. How can I express $y_2$ in function of $D$ and $x$?
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$\begingroup$ As far as I remember, it should be enough to take the conjugate of D and conjugate of x.(Have you tried to google the rules of complex conjugate?) $\endgroup$– IrreducibleJun 29, 2022 at 5:50
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1 Answer
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You just need to conjugate the matrix $D$ as well as the vector $x$, and flip the rows of $D$ upside down, i.e., the first row becomes the last row, etc. I.e., you need to introduce a new matrix $\tilde{D}$ with the rows of $D$ flipped:
$$y_2=\tilde{D}^*x^*$$
where $^*$ denotes complex conjugation.
