Eigenvalues of certain matrices

These are some interesting matrices. I did not find that $B^n=I_{2d}$ always, though this was true when $n$ was a multiple of $d$. The case of $r=1$ seemed different from the more general case $r>1$ and I did not consider it in detail since it may not be of interest.

Preliminaries

The matrices $R(\theta ),~\theta \in \mathbb{R}$ comprise a family of mutually commuting orthogonal matrices with common eigenvectors. The unitary matrix $U$ with columns of the normalized eigenvectors can be used to diagonalize any $R(\theta )$, $ R(\theta )=U\Lambda U^{\ast },$ $ \Lambda =U^{\ast }R(\theta )U$, where

$$ \Lambda = \begin{bmatrix} \exp (+2\pi i\theta ) & 0 \\ 0 & \exp (-2\pi i\theta ) \end{bmatrix} ,~U=\frac{1}{\sqrt{2}} \begin{bmatrix} -i & i \\ 1 & 1 \end{bmatrix} \text{.} $$

Therefore any product of the $2\times 2$ blocks here can be easily diagonalized, $$U^{\ast }R(\theta )R(\theta ^{\prime })U=U^{\ast }R(\theta )UU^{\ast }R(\theta ^{\prime })U=\Lambda \Lambda ^{\prime }$$, which significantly simplifies the problem.

Let $R_{m}=R(1/m)$, then the powers $R_{m}^{k}=R(k/m)$ have eigenvalues $ \exp (\pm 2\pi ik/m)$. $R_{m}^{k}$ is equal to $I_{2}$ when $k=m$, i.e., the sequence $R_{m}^{k},~k=0,1,\ldots $has period $m$. More generally the sequence $\left( R_{m}^{j}\right) ^{k},~k=0,1,\ldots $ has period $\operatorname{lcm} (j,m)/j\leq m$.

Eigenvalues of $A^{a}$

The eigenvalues of $A$ are the union of the eigenvalues of its blocks, and therefore are $\exp (\pm 2\pi ir^{k}/m),k=0,\ldots ,d-1$. The eigenvalues of $A^{a}$ are $\exp (\pm 2\pi ir^{k}a/m),k=0,\ldots ,d-1,$ and these are all unity for $a=m$, i.e., $A^{m}=I_{2d}$ and the sequence of the $A^{a}$ has period $m$. Since the first block of $A$ is $R_{m}$ with period $m$, there cannot be a shorter period for the $A^{a}$.

Period of $B^{b}$

The situation for $B$ is more complicated. The block $R_{n}^{d}$ has period $ p:=\operatorname{lcm}(d,n)/d\leq n$, but the block structure is not diagonal and has its own period. It is instructive to write the first few matrices in the sequence $B^{b},b=0,1,\ldots $ for the case $d=4$. ($R_{n}^{d}$ is here abbrieviated as $R$ and $I_{2}$ as $I$.)

$$ \begin{bmatrix} I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & 0 & 0 & I% \end{bmatrix}% ,~ \begin{bmatrix} 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & 0 & 0 & I \\ R & 0 & 0 & 0 \end{bmatrix} ,~ \begin{bmatrix} 0 & 0 & I & 0 \\ 0 & 0 & 0 & I \\ R & 0 & 0 & 0 \\ 0 & R & 0 & 0 \end{bmatrix} ,~ \begin{bmatrix} 0 & 0 & 0 & I \\ R & 0 & 0 & 0 \\ 0 & R & 0 & 0 \\ 0 & 0 & R & 0 \end{bmatrix} , $$

$$ \begin{bmatrix} R & 0 & 0 & 0 \\ 0 & R & 0 & 0 \\ 0 & 0 & R & 0 \\ 0 & 0 & 0 & R \end{bmatrix} ,~% \begin{bmatrix} 0 & R & 0 & 0 \\ 0 & 0 & R & 0 \\ 0 & 0 & 0 & R \\ R^{2} & 0 & 0 & 0 \end{bmatrix} ,~% \begin{bmatrix} 0 & 0 & R & 0 \\ 0 & 0 & 0 & R \\ R^{2} & 0 & 0 & 0 \\ 0 & R^{2} & 0 & 0 \end{bmatrix} ,\ldots $$

The zero/non-zero block structure has periodicity $d$: $B$ becomes block diagonal when $b$ is a multiple of $d$. Each such quasi-period gains an additional factor of $D_{R}=\operatorname{diag}(R,R,\ldots )$ and the true period occurs when $D_{R}^{p}=I_{2d}$ and $R^{p}=I$. That is, the period of $ B^{b}$ will be $\operatorname{lcm}(p,d)$. If $n$ is a multiple of $d$ then $ B^{n}=I_{2d}$

Eigenvalues of $B^{b}$

The matrix $B$ is in block companion matrix form, with the $2\times 2$ blocks along the last row corresponding to the negated coefficients of the characteristic matrix polynomial $I_{2}x^{d}-R_{n}^{d}$ .Applying the unitary similarity with $U$ gives $I_{2}x^{d}-\operatorname{diag}(\rho ,\overline{ \rho })$, where $\rho $ and $\overline{\rho }$ are the eigenvalues of $ R_{n}^{d}$. Taking the determinant gives $(x^{d}-\rho )(x^{d}-\overline{\rho })=0$ and solving for $x$ gives the $2d$ eigenvalues for $B,$ namely $\omega ^{k}\rho ^{1/d}$ and $\omega ^{k}\overline{\rho }^{1/d}$, $k=0,\ldots ,d-1$ where $\omega $ is the principal $d$th root of unity. Explicitly, these are $ \exp (2\pi ik/d)\exp (\pm 2\pi i/n),$ $k=0,\ldots ,d-1$. It follows that the eigenvalues of $B^{b}$ are $\exp (2\pi ikb/d)\exp (\pm 2\pi ib/n),$ $ k=0,\ldots ,d-1$. For the case $b\operatorname{mod}d=0,$ the matrices $B^{b}$ are block diagonal and the first factor simplifies to $1$; the eigenvalues are then $d$ copies of a complex-conjugate pair.

Eigenvalues of $A^{a}B^{b}$

We first show that the $2d\times 2d$ problem splits into two $d\times d$ problems. Note that $B^{b}$ has one block in each row and column, and premultiplying by the block diagonal $A^{a}$ multiplies the block in the $i$% th pair of rows of $B^{b}$ by the $i$th block of $A^{a}$, preserving the block structure. Performing the similarity $U_{2d}^{\ast }A^{a}B^{b}U_{2d}$ with $U_{2d}=\operatorname{diag}(U,U,\ldots )$ diagonalizes all the blocks, with the diagonal entries of each block now containing the products of the respective eigenvalues of the matrix factors in the block.

A permutation similarity that moves odd rows and columns to the beginning and even rows and columns to the end generates a block diagonal form with the first $% d\times d$ block, $W$, having the same structure as $A^{a}B^{b}$ but having the first eigenvalues from each $2\times 2$ block, and with the second $% d\times d$ block having the second eigenvalues. Therefore the eigenvalues of $A^{a}B^{b}$ come in two sets of $d$, with the second being complex conjugates of the first.

We deal now only with the first $d\times d$ block. We write $\alpha _{1},\ldots ,\alpha _{d}$ for the first eigenvalues of each $2\times 2$ block of $A,$ $\lambda _{i}=\alpha _{i}^{a}$ for the first eigenvalues of the blocks of $A^{a}$, and (as above) $\rho $ for the first eigenvalue of $R_{n}^{d}$. For example, with $d=4$ and $b=5$ we have

$$ W= \begin{bmatrix} 0 & -\alpha _{1}^{a}\rho & 0 & 0 \\ 0 & 0 & -\alpha _{2}^{a}\rho & 0 \\ 0 & 0 & 0 & -\alpha _{3}^{a}\rho \\ -\alpha _{4}^{a}\rho ^{2} & 0 & 0 & 0 \end{bmatrix} $$ $$ =\operatorname{diag}\left( \lambda _{1},\ldots \lambda _{4}\right) \operatorname{diag} \left( \rho ,\rho ,\rho ,\rho ^{2}\right) \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} $$

where the last matrix is $C_{4}^{5}=C_{4}$, the basic circulant permutation matrix of order $4$, corresponding to the permutation $\sigma =(1234)$. In the common case, as here, where the permutation has only one cycle, the characteristic polynomial $\left\vert xI_{d}-W\right\vert $ has only one term other than $x^{d}$, which arises from the product of all the nonzero entries, with sign $(-1)^{d}\operatorname{sgn}\sigma =-1$. The characteristic polynomial is

$$ x^{d}-\rho ^{b}\lambda _{1}\lambda _{2},...,\lambda _{d} $$

so its roots are the eigenvalues

$$ \left( \rho ^{b}\prod_{i}\lambda _{i}\right) ^{1/d}\omega ^{k},~k=0,1,\ldots ,d-1. $$

where the $\omega ^{k}\ $are the $d$th roots of unity. Using the general theory below, this case occurs when $d$ and $b$ are coprime.

The product $\prod_{i}\lambda _{i}$ of first eigenvalues of the blocks of $A^{a}$ can be evaluated using the formula above as

$$ \prod_{k=0}^{d-1}\exp (+2\pi ir^{k}a/m)=\exp \left( +2\pi i(a/m)\sum_{k=0}^{d-1}r^{k}\right) =\exp \left( +2\pi i\frac{a\left( r^{d}-1\right) }{m\left( r-1\right) }\right) =1 $$

where the last equality uses $r^{p}\operatorname{mod}m=1$ or $m|r^{p}-1$. For the second eigenvalues replace the explicit $+$ with $-$; the result is the same. (For $r=1,$ $\sum_{k=0}^{d-1}r^{k}=d$ and the relationship may not hold.) That is, the eigenvalues of $A^{a}B^{b}$ are the same as those of $% B^{b}$ for the common case of a single-cycle permutation.

Another simple case is when $b\operatorname{mod}d=0$. This is the case where $W$ is diagonal, $C_{d}=I_{d}$, and $B^{b}$ is in block diagonal form. The permutation has only cycles of length 1 (fixed points). The blocks of $B^{b}$ are all $\left( R_{n}^{d}\right) ^{b/d}=R_{n}^{b}$ with eigenvalues $\exp (\pm 2\pi ib/n)$ and combine with the respective block sof $A^{a}$ so the eigenvalues of $A^{a}B^{b}$ are the respective products

$$ \exp (\pm 2\pi ir^{k}a/m)\exp (\pm 2\pi ib/n),k=0..d-1 $$

Here both first and second eigenvalues are included, with upper and lower signs respectively.

In the general case where the permutation has multiple cycles, the characteristic polynomial factors in a pattern related to the cycle lengths. Then some eigenvalues have products of only a proper subset of $\lambda _{i}$ and the eigenvalues of $A^{a}B^{b}$ generally will not be the same as those of $B^{b}$. The general form is $W=D_{A^{a}}D_{\rho ,b}C_{d}^{b}$, where $% D_{A^{a}}=\operatorname{diag}\left( \lambda _{1},\ldots \lambda _{d}\right) $ and

$$ D_{\rho ,b}=\operatorname{diag}\left( \rho ^{\lfloor b/d\rfloor },\ldots ,[d-(d \operatorname{mod}b)\,\mathrm{times}],\rho ^{\lfloor b/d\rfloor +1},\ldots ,[d\operatorname{ mod}b~\mathrm{times}]\right) . $$

The eigenvalues for the product of a diagonal matrix and a permutation matrix (a PD or monomial matrix) are discussed in Ref. 1. Working this through for the case here leads to the following prescription. Find the cycles in the permutation $(1,2,\ldots ,d)^{b}$. There are $s=\gcd (d,b)$ cycles each of length $l=d/s$. For each cycle of length $l$, there is a set of $l$ first eigenvalues given by

$$ \left( \rho ^{b/s}\prod_{cycle}\lambda _{i}\right) ^{1/l}\omega _{l}^{k},~k=0,1,\ldots ,l-1. $$

where the $\prod_{cycle}\lambda _{i}$ means the product of $ \lambda _{i}$ with subscripts that are the $l$\ numbers in the cycle, and $ \omega _{l}=\exp (2\pi i/l)$. As before the second eigenvalues are the complex conjugates of these.

1. P.J. Davis, Circulant Matrices, Wiley, 1979, Sec 5.3, p. 166.