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We have a block matrix:

$$ \left[\begin{array}{c|c|c} A & 0 & 0 \\ \hline 0 & B & 0 \\ \hline 0 & 0 & C \end{array}\right] $$

Here $A$, $B$ and $C$ are all permutation matrices of varying sizes, raised to a power. For example, all of the block matrices take on forms such as:

$$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{bmatrix}^a, \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix}^b, \text{ or } \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 & 0\\ \end{bmatrix}^c, $$

We can suppose that we can only obtain one copy of this matrix. In other words, we are not allowed to use this block matrix more than once. Can we assign corresponding scalars to the matrices $(A \to a, B \to b, C \to c)$ and (using matrix multiplication on the block matrix) get a result $a \cdot b \cdot c$?

WHAT I MEAN BY ASSIGNING SCALARS

Suppose that $A$ is only one of the following:

$$ \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \text{or} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$

...and $B$ is one of the following:

$$ \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \text{or} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$

We could say that the first $A$ matrix is equivalent to $1$, and the second $A$ matrix is equivalent to 2. Also, the first $B$ equivalent to 3 and the second to 4. Then if we started with the first $A$ and the second $B$, we would be looking to find a result of $1 \cdot 4$. If we started with the second $A$ and the second $B$ we would be trying to obtain a result $2 \cdot 4$.

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  • $\begingroup$ What do you mean by "assign a scalar to a matrix"? $\endgroup$ – Jim Sep 6 '13 at 17:41
  • $\begingroup$ or by "obtain one copy" of a matrix? $\endgroup$ – Jim Sep 6 '13 at 17:42
  • $\begingroup$ @Jim: I've tried to clarify things in the question. I hope this helps. $\endgroup$ – Matt Groff Sep 6 '13 at 17:50
  • $\begingroup$ Why not just take diagonal matrices then? If $I$ is the $2 \times 2$ identity then $(aI)\cdot(bI)\cdot(cI) = abcI$. $\endgroup$ – Jim Sep 6 '13 at 17:59
  • $\begingroup$ @Jim: The problem is that I can't always get the block matrices as diagonal matrices. However, there may be a transformation that transforms the block matrices into diagonal matrices. That would probably solve the question. To restate, I can't always start with diagonal matrices, so that is part of the problem. $\endgroup$ – Matt Groff Sep 6 '13 at 18:02

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