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A $2k\times 2k$ circulant matrix $\ C$ takes the form

\begin{align} C= \begin{bmatrix} c_0 & c_{2k-1} & \dots & c_{2} & c_{1} \\ c_{1} & c_0 & c_{2k-1} & & c_{2} \\ \vdots & c_{1}& c_0 & \ddots & \vdots \\ c_{2k-2} & & \ddots & \ddots & c_{2k-1} \\ c_{2k-1} & c_{2k-2} & \dots & c_{1} & c_0 \\ \end{bmatrix}. \end{align}

Now we consider a $2k\times 2k$ generalized circulant matrix $\ C_{1}$

\begin{align} C_{1}= \begin{bmatrix} c_0 & -c_{2k-1} & \dots & c_{2} & -c_{1} \\ c_{1} & -c_0 & c_{2k-1} & & -c_{2} \\ \vdots & -c_{1}& c_0 & \ddots & \vdots \\ c_{2k-2} & & \ddots & \ddots & -c_{2k-1} \\ c_{2k-1} & -c_{2k-2} & \dots & c_{1} & -c_0 \\ \end{bmatrix}. \end{align}

That is the first column shifts with a multiplication of $-1$.

Then what are the eigenvalues of $\ C_{1}$? Thank you!

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$C_1$ isn't quite a "generalized $2k \times 2k$ circulant", because it does not include $C$ as a special case. It is, however, block-circulant: let $A_0,\ldots,A_{k-1}$ be the $2\times 2$ matrices $$ A_j = \left[\begin{array}{cc} c_{2j} & -c_{2j-1} \\ c_{2j+1} & -c_{2j} \end{array}\right] \quad(j=0,1,\ldots,k-1) $$ (with $c_{-1}$ interpreted as $c_{2k-1}$); then $C_1$ is a $k\times k$ block-circulant with first column $A_0,\ldots,A_{k-1}$. As with ordinary circulant matrices, a block-circulant matrix is conjugate over $\bf C$ to a block-diagonal matrix with $n$-th block $$ \hat A_n := \sum_{j=0}^{k-1} e^{2\pi i nj/k} A_j. $$ So the eigenvalues of $C_1$ are the eigenvalues of the $\hat A_n$, which in this $2 \times 2$ case are given by the formula $\frac12(t \pm \sqrt{t^2-4\Delta})$ where $t$ is the trace and $\Delta$ is the determinant.

In the present case, each $A_j$ has trace zero, so the same is true of all the $\hat A_n$, and the eigenvalues are simply $\pm \sqrt{-\det \hat A_n}$. This suggests an alternative description. Form the matrix $C_1^2$, and note that its entries are zero when the row and column index have opposite parity. Separating odd and even indices we find that $C_1^2$ is conjugate with a block-diagonal matrix $\left[{C \; 0 \atop 0 \; C}\right]$ for some $k \times k$ matrix $C$ that is circulant in the usual sense. Compute the eigenvalues of this circulant matrix $C$ as usual, and then extract their square roots to recover the eigenvalues of $C_1$.

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  • $\begingroup$ could you please clarify what "conjugate over $\mathbb{C}$ means? Is it a similarity of the form $A=P^H B P$? $\endgroup$ – kodlu Nov 29 '17 at 10:59

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