# Trace of Product of Powers of $A$ and $A^\ast$

Let $n$ be odd, $\displaystyle v=1,...,\frac{n-1}{2}$ and $\displaystyle \zeta=e^{2\pi i/n}$.

Define the following matrices:

$$A(0,v)=\left(\begin{array}{cc}1+\zeta^{-v} & \zeta^v+\zeta^{2v}\\ \zeta^{-v}+\zeta^{-2v}&1+\zeta^{v}\end{array}\right),$$ $$A(1,v)=\left(\begin{array}{cc}\zeta^{-1}+\zeta^{-v} & \zeta^{v}\\ \zeta^{-v}&\zeta^{-1}+\zeta^{v}\end{array}\right).$$ $$A(n-1,v)=\left(\begin{array}{cc}\zeta+\zeta^{-v} & \zeta^{2v}\\ \zeta^{-2v}&\zeta+\zeta^v\end{array}\right).$$

I am hoping to calculate for each of these $A$ $$\text{Tr}\left[\left(A^k\right)^*A^k\right]=\text{Tr}\left[\left(A^*\right)^kA^k\right].$$

All I have is that $A$ and $A^*$ in general do not commute so I can't simultaneously diagonalise them necessarily.

I do know that if we write $A=D+(A-D)$ (with $D$ diagonal), that $$A^*=\overline{D}+(A-D).$$

I suppose anybody who knows anything about calculating $$\text{Tr}(A^kB^k)$$ can help.

Context: I need to calculate or rather bound these traces to calculate a distance to random for the convolution powers of a $\nu\in M_p(\mathbb{G}_n)$ for $\mathbb{G}_n$ a series of quantum groups of dimension $2n^2$ ($n$ odd). For $u=2,...,k-2$, $A(u,v)$ is diagonal so no problems there.

• Are you sure about $\operatorname{Tr}(A) = \operatorname{Tr}(A^*)$? Seems to be not true at least for $A(1,v)$? – Calle Nov 9 '15 at 21:48
• @Calle hmmm... I am not sure if I have a typo or not --- I am away from my work. I will get back to you in the morning. Perhaps they are wrong as given and they should be $\zeta+\zeta^v$ and $\zeta^{-1}+\zeta^{-v}$ in $[A(1,v)]_{22}$ and $[A(n-1,v)]_{11}$... I might just throw that in for now... – JP McCarthy Nov 9 '15 at 22:01
• That makes more sense. :) – Calle Nov 9 '15 at 22:17
• @Calle --- I am not 100% on that but sure we'll see. – JP McCarthy Nov 9 '15 at 22:22
• Sorry, maybe I do not understand the notation: isn't it just that $(A^*)^k=(A^k)^*$ ? – Start wearing purple Nov 9 '15 at 22:26

The first case is easy. Let $A:=A(0,v)$ and write $$A= \left(1+\zeta^v\right)\left(\begin{array}{cc}\zeta^{-v} & \zeta^v \\ \zeta^{-2v} & 1\end{array}\right)=\left(1+\zeta^v\right) \alpha^T\otimes \beta ,$$ where $\alpha=\left(1\;\;\zeta^{-v}\right)$, $\beta=\left(\zeta^{-v}\;\; \zeta^v\right)$. This implies that $$A^*=\left(1+\zeta^{-v}\right)\bar{\beta}^T\otimes\bar{\alpha}.$$ That both matrices have rank $1$ reduces the computation of traces to scalar products of $\alpha,\bar{\alpha},\beta,\bar{\beta}$. One has for instance \begin{align} \operatorname{Tr}\left(\left(A^*\right)^kA^k\right)&= \left(1+\zeta^v\right)^k\left(1+\zeta^{-v}\right)^k \left(\bar\alpha \cdot \bar{\beta}^T\right)^{k-1}\left(\bar{\alpha}\cdot \alpha^T\right)\left(\beta\cdot \alpha^T\right)^{k-1}\left(\beta\cdot\bar{\beta}^T\right)=\\ &=4\left(1+\zeta^v\right)^{2k-1}\left(1+\zeta^{-v}\right)^{2k-1}. \end{align}

In the other two cases, I do not see a clever method, but a straightforward approach would work as well. Diagonalize $A,A^*$ as $$A=PDP^{-1},\qquad A^*=P^{-*}\bar{D}P^*,$$ then $$\operatorname{Tr}\left(\left(A^*\right)^kA^k\right)= \operatorname{Tr}\left(PD^kP^{-1}P^{-*}\bar{D}^kP^*\right),$$ with $D$, $\bar{D}$ diagonal. Thus one only needs to compute diagonalizing transformation $P$ built from the eigenvectors of $A$ and then to compute the trace of the product of six matrices.

• I changed $A(1,v)$ and $A(n-1,v)$. Are they anymore amenable to your first method? – JP McCarthy Nov 10 '15 at 8:32
• ...although I don't see any problem with the second method... I will work through and probably award the bounty to you then. – JP McCarthy Nov 10 '15 at 8:49
• @JpMcCarthy For the first method to work, $\operatorname{det} A$ should vanish, which is not the case for the other two matrices. – Start wearing purple Nov 10 '15 at 9:16
• Yeah this is working lovely thank you. – JP McCarthy Nov 10 '15 at 13:49
• @JpMcCarthy You are welcome! – Start wearing purple Nov 10 '15 at 14:02

## Solution for $A_0 = A(0,v)$

\begin{align*} A_0 &= \begin{pmatrix} 1+\zeta^{-v} & \zeta^v + \zeta^{2v} \\ \zeta^{-v}+\zeta^{-2v} & 1+\zeta^v \end{pmatrix} = (\zeta^{-\frac{1}{2}v}+\zeta^{+\frac{1}{2}v}) \begin{pmatrix} \zeta^{-\frac{1}{2}v} & \zeta^{\frac{3}{2}v} \\ \zeta^{-\frac{3}{2}v} & \zeta^{\frac{1}{2}v} \end{pmatrix} \\ &=: \lambda \hat{A}_0 \\ \hat{A}^2_0 &= \begin{pmatrix} \zeta^{-\frac{1}{2}v} & \zeta^{\frac{3}{2}v} \\ \zeta^{-\frac{3}{2}v} & \zeta^{\frac{1}{2}v} \end{pmatrix}^2 = \begin{pmatrix} 1+\zeta^{-v} & \zeta^v + \zeta^{2v} \\ \zeta^{-v}+\zeta^{-2v} & 1+\zeta^v \end{pmatrix} \\ &= A_0 \\ \implies A_0^2 &= \lambda^2 \hat{A}_0^2 = \lambda^2 A_0 \\ \implies A_0^k &= \lambda^{2(k-1)} A_0 \end{align*} Moreover note that $\lambda$ is real. In fact, $\lambda=2\cos(\frac{v}{n}\pi)$. Therefore: \begin{equation*} {\tt tr}((A_0^*)^kA_0^k)) = \lambda^{4(k-1)}{\tt tr}(A_0^* A_0) = 4\lambda^{4(k-1)} \end{equation*} Note that $\lambda^2 = (\zeta^{-\frac{1}{2}v}+\zeta^{+\frac{1}{2}v})^2 = (1+\zeta^v) (1+\zeta^{-v})$, hence the solution is equivalent to one by Start wearing purple.

## Solution for $A_1 = A(1,v)$

\begin{align*} A_1 &= \begin{pmatrix} \zeta^{-1}+\zeta^{-v}& \zeta^v \\ \zeta^{-v} & \zeta^{-1}+\zeta^v \end{pmatrix} = \zeta^{-1}\begin{pmatrix} 1&0\\0&1 \end{pmatrix} + \begin{pmatrix} \zeta^{-v} & \zeta^{v} \\ \zeta^{-v} & \zeta^{v} \end{pmatrix} \\ &=: \zeta^{-1} I + \hat{A}_1 \\ \hat{A}_1^2 &= \begin{pmatrix} \zeta^{-2v}+1 & \zeta^{2v}+1 \\ \zeta^{-2v}+1 & \zeta^{2v}+1 \end{pmatrix} = (\zeta^{-v} + \zeta^v) \begin{pmatrix} \zeta^{-v} & \zeta^v \\ \zeta^{-v} & \zeta^v \end{pmatrix} \\ &=: \mu \hat{A}_1 \\ \implies \hat{A}_1^k &= \mu^{k-1} \hat{A}_1 \\ \implies A_1^k &= \big(\zeta^{-1}I + \hat{A}_1\big)^k = \sum_{j=0}^{k}\binom{k}{j} \zeta^{j-k}\hat{A}_1^j = \zeta^{-k}\bigg(I + \hat{A}_1 \sum_{j=1}^{k}\binom{k}{j} \zeta^{j}\mu^{j-1}\bigg) \\&= \zeta^{-k}\bigg(I + \frac{1}{\mu}\hat{A}_1\bigg[-1 + \sum_{j=0}^{k}\binom{k}{j} \zeta^{j}\mu^{j}\bigg] \bigg) \\&= \zeta^{-k}\bigg(I + \frac{(1-\zeta\mu)^k - 1}{\mu}\hat{A}_1\bigg) \end{align*} Here we again have that $\mu=2\cos(2\pi\frac{v}{n})$ is real; moreover we have ${\tt tr}(\hat{A}_1^*) = {\tt tr}(\hat{A}_1) = \mu$ and ${\tt tr}(\hat{A}_1^*\hat{A}_1)=4$. Hence if we let $\alpha_k = (1-\zeta\mu)^k - 1$ the trace is \begin{equation*} {\tt tr}((A_1^*)^kA_1^k)) = 1 + \alpha_k + \bar\alpha_k + \frac{4}{\mu^2}\alpha_k\bar\alpha_k \end{equation*}

## Solution for $A_{n-1} = A(n-1,v)$

\begin{align*} A_{n-1} &= \begin{pmatrix} \zeta+\zeta^{-v}& \zeta^{2v} \\ \zeta^{-2v} & \zeta + \zeta^{v} \end{pmatrix} = \zeta \begin{pmatrix} 1&0\\0&1 \end{pmatrix} + \begin{pmatrix} \zeta^{-v}& \zeta^{2v} \\ \zeta^{-2v} & \zeta^{v} \end{pmatrix} \\&=: \zeta I + \hat{A}_{n-1} \\ \hat{A}_{n-1}^2 &= \begin{pmatrix} \zeta^{-2v} + 1 & \zeta^{v}+\zeta^{3v} \\ \zeta^{-3v} + \zeta^{-v} & \zeta^{2v}+1 \end{pmatrix} = (\zeta^{-v} + \zeta^{v}) \begin{pmatrix} \zeta^{-v}& \zeta^{2v} \\ \zeta^{-2v} & \zeta^{v} \end{pmatrix} \\&=: \mu \hat{A}_{n-1} \\ \implies \hat{A}_{n-1}^k &= \mu^{k-1}\hat{A}_{n-1} \\ \implies A_{n-1}^k &= (\zeta I + \hat{A}_{n-1})^k = \ldots\\ &= \zeta^{k}\bigg(I + \frac{(1-\zeta^{-1}\mu)^k - 1}{\mu}\hat{A}_{n-1}\bigg) \end{align*} Here we immediately observe that this is remarkably similar to the previous solution. In fact, since ${\tt tr}(\hat{A}_{n-1}) = \mu$ and ${\tt tr}(\hat{A}_{n-1}^*\hat{A}_{n-1})=4$ aswell, the traces are the same!! \begin{equation*} {\tt tr}((A_{n-1}^*)^kA_{n-1}^k)) = 1 + \alpha_k + \bar\alpha_k + \frac{4}{\mu^2}\alpha_k\bar\alpha_k = {\tt tr}((A_1^*)^kA_1^k)) \end{equation*}