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I am trying to code this problem: Suppose there are four elements to use when tensoring: $$ \left\{ I = \begin{bmatrix}1&0\\0&1\end{bmatrix},\, X = \begin{bmatrix}0&1\\1&0\end{bmatrix},\, Y = \begin{bmatrix}0&-i\\i&0\end{bmatrix},\, Z = \begin{bmatrix}1&0\\0&-1\end{bmatrix} \right\}, $$

where $i = \sqrt{-1}$.

The tensor product is denoted by $\otimes$ and applied on two matrices $A_{n \times n}$ and $B_{m \times m}$ as such: $$ A\otimes B = \begin{align} \begin{bmatrix} A_{11}B & A_{12}B & \ldots & A_{1n}B\\ A_{21}B & A_{22}B & \ldots & A_{2n}B\\ \vdots & \vdots & \ddots & \vdots \\ A_{n1}B & A_{n2}B & \ldots & A_{nn}B\\ \end{bmatrix}_{nm\times nm} \end{align} $$

You are given the following: $$ \begin{align} A_1 &= a_{11} \otimes a_{12} \otimes \cdots \otimes a_{1n}\\ A_2 &= a_{21} \otimes a_{22} \otimes \cdots \otimes a_{2n}\\ &\phantom{~~}\vdots\\ A_m &= a_{n1} \otimes a_{m2} \otimes \cdots \otimes a_{mn}\\ \end{align}, $$ where $a_{ij} \in \{I, X, Y, Z\}$.

Let $A = A_1 * A_2 * \cdots * A_m$, where $*$ is ordinary left-to-right matrix multiplication. I would assume there is some tensor product combination that equals $A$, i.e., $$ A = a_{1} \otimes a_{2} \otimes \cdots \otimes a_{n}, \text{ where } a_i \in \{I,\, X,\, Y,\, Z\}.\\ $$ How can I find the tensor product combination of $A$? Please keep in mind that I don't know if my assumption is correct. Furthermore, there are similar questions that have been asked such as "Inverse" of tensor product and The Inverse of a Fourth Order Tensor but the details and requirements of my question differ.

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  • $\begingroup$ What do you mean by tensor product of matrices? For instance, what would $I\otimes X$ be? $\endgroup$
    – Berci
    Commented Sep 4, 2021 at 8:02
  • $\begingroup$ @Berci, I have edited the question! As for your request:$ I \otimes X = \begin{bmatrix} 1\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}& 0\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\\ 0\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}& 1\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\\ \end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{bmatrix} $ $\endgroup$ Commented Sep 4, 2021 at 8:44
  • $\begingroup$ Ok. But then the sizes of $A_i$ and $A$ might differ.. How could they be ever equal? $\endgroup$
    – Berci
    Commented Sep 4, 2021 at 8:48
  • $\begingroup$ The sizes of $A_i$ are the same (say they are $n \times n$ for simplicity). The size of $A$ is also $n \times n$ because it is the matrix multiplication (not tensoring) of all $A_i$s. $\endgroup$ Commented Sep 4, 2021 at 8:52
  • $\begingroup$ This fails even in the case where $n=1$; just take $A_1=a_{11}=X$, $A_2=a_{21}=Y$ then $A_1 A_2$ falls outside your set. Surely you need to extend the set to be the full quaternion group? $\endgroup$ Commented Sep 4, 2021 at 14:03

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