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Start Wearing Purple answered this question for me. I am now asking a question about their answer as I can make the question general enough to be useful on it's own.

For $1\times 2$ matrices $\alpha,\,\beta\in M_{1\times 2}(\mathbb{C})$, SWP uses the following equality

$$\operatorname{Tr}\left[(\overline{\alpha}\otimes\beta^*)^k(\alpha^T\otimes \beta)^k\right]=(\bar{\alpha}\cdot \beta^*)^{k-1}(\overline{\alpha}\cdot \alpha^T)(\beta\cdot \alpha^T)^{k-1}(\beta\cdot \beta^*).\,\,\,\,\,\,(\star)$$ where the dots are scalar products/matrix multiplications.

To be honest I don't really have a clue how to deal with these having never used Kronecker products before. I have used the result in my work but do not yet understand the above calculation.

I am looking for somebody to explain this calculation $(\star)$ to me or perhaps send me to an appropriate online reference.

There is a chance that the exact nature of $\alpha$ and $\beta$ make this calculation true:

$$\alpha:=(1\,\,\,\,\,\zeta^{-v})\,\,\,\text{ and }\,\,\,\beta:=(\zeta^{-v}\,\,\,\,\, \zeta^v),$$ for $\zeta:=e^{2\pi i/n}$.

Context: I used this result to help bound the number of steps required for a random walk on the Sekine quantum groups to converge to the Haar state. More background here.

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We will need a few things: \begin{align} \text{ conjugate-rule:}&&& (A\otimes B )^*= A^* \otimes B^* &\text{note they don't change places!} \\ \text{ mixed-product-rule:}&&& (A\otimes B )(C \otimes D) = (AC)\otimes (BD) &\text{if matrices have correct dim} \\ \text{ trace-rule:}&&& {\tt tr}(A \otimes B) = {\tt tr}(A) {\tt tr}(B) \\ \text{ trace-product:}&&& {\tt tr}(A^* B) = {\tt tr}(A B^*) \\ \text{ trace-linearity:}&&& {\tt tr}(\lambda A) = \lambda {\tt tr}(a) &\text{for scalar $\lambda$} \end{align}

Now note that since $\alpha$ and $\beta$ are $1\times 2$, $${\tt (I)}\qquad (\alpha^T\otimes \beta)^* (\alpha^T\otimes \beta) = (\bar\alpha\otimes\beta^*)(\alpha^T\otimes \beta) = (\underbrace{\bar\alpha\alpha^T}_{1\times 1} \otimes \underbrace{\beta^*\beta}_{2\times 2}) = (\bar\alpha\alpha^T)(I_1 \otimes \beta^*\beta) $$

And now we on the one hand: $${\tt (II)}\qquad (\alpha^T\otimes \beta)^* (I_1 \otimes \beta^*\beta) = (\bar\alpha\otimes \beta^*) (\beta^*\beta \otimes I_1) = ((\underbrace{\bar\alpha \beta^*}_{1 \times 1})\beta \otimes \beta^*) = (\bar\alpha \beta^*)(\beta \otimes \beta^*) $$ and on the other $$ {\tt (III)}\qquad (\beta \otimes \beta^*)(\alpha^T\otimes \beta) = (\underbrace{\beta\alpha^T}_{1\times 1}\otimes \beta^*\beta) = (\beta\alpha^T)(I_1 \otimes \beta^*\beta) $$ So Combining (I), (II) and (III) we find that by sort of 'ping-ponging' the middle Kroner-product back and forth that we can extract more and more scalar terms by reducing the powers:

\begin{align} \big((\alpha^T\otimes \beta)^*\big)^k (\alpha^T\otimes \beta)^k &= (\bar\alpha\alpha^T) \big((\alpha^T\otimes \beta)^*\big)^{k-1} (I_1 \otimes \beta^*\beta)(\alpha^T\otimes \beta)^k \\&= (\bar\alpha\alpha^T)(\bar\alpha \beta^*)^{k-1}(\beta\alpha^T)^{k-1}(I_1 \otimes \beta^*\beta) \\&= (\bar\alpha\alpha^T)(\bar\alpha \beta^*)^{k-1}(\beta\alpha^T)^{k-1}(\beta^* \beta) \end{align}

From which your identity can be immediately deduced by the trace linearity and product rule: $$ {\tt tr}(\underbrace{\beta^*\beta}_{2x2}) = {\tt tr}(\underbrace{\beta\beta^*}_{1 \times 1}) = \beta\beta^*$$

Note that since $\beta$ is a row-vector holds: $(I_1 \otimes \beta^*\beta) = (\beta^*\beta \otimes I_1) = \beta^*\beta =(\beta^*\otimes \beta )$. In fact all of the above calculations immediately generalize to the case of $1\times M$ vectors.

If you want to I can also post an answer how to obtain the equality $${\tt tr}((A^*)^k A^k) = 4\left(1+\zeta^v\right)^{2k-1}\left(1+\zeta^{-v}\right)^{2k-1}$$ from your original post without the need of using Kronecker products.

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  • $\begingroup$ Thank you I will have a proper look on Monday and presumably award you the bounty. $\endgroup$ – JP McCarthy Nov 18 '16 at 23:04
  • $\begingroup$ Thank you very much. I will award the bounty. I would also be most interested in seeing your answer to the original question. I have put up a bounty of 400 on that so that I could reward you for it. $\endgroup$ – JP McCarthy Nov 21 '16 at 12:03
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    $\begingroup$ It's really just the observation that $$ \begin{pmatrix}\zeta^{-v} & \zeta^v \\ \zeta^{-2v} & 1\end{pmatrix}^2 = \begin{pmatrix} \zeta^{-2v} + \zeta^{-v} & 1 + \zeta^{v} \\ \zeta^{-3v} + \zeta^{-2v} & \zeta^{-v} + 1 \end{pmatrix} = (1+\zeta^{-v}) \begin{pmatrix}\zeta^{-v} & \zeta^v \\ \zeta^{-2v} & 1\end{pmatrix} $$ $\endgroup$ – Hyperplane Nov 21 '16 at 12:14
  • $\begingroup$ I have the bounty up anyway if you want to expand that into a full answer on the original question... $\endgroup$ – JP McCarthy Nov 21 '16 at 12:39

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