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I am hazarding a guess, I believe this should do the job A = { {1, x, 0}, {0, -1, 0}, {0, y, 1} } It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols B = { {0, 0, -$i$}, {$i$, 0, 0}, {0, 1, 0} } Then A.A= B.B.B = Id the order of A.B should $A.B$ would be infinite as when $x$ and $y$ are suitably chosen, for example one can choose $x$ and $y$ so that the coefficient of the matrices are $(A.B)^n$ unbounded ? |
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I am hazarding a guess, I believe this should do the job A = { {1, x, 0}, {0, -1, 0}, {0, y, 1} } It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols B = { {0, 0, -\iota}, $i$}, {\iota, $i$, 0, 0}, {0, 1, 0} } Then A.A= B.B.B = Id the order of A.B should be infinite as the coefficient of the matrices are unbounded ? |
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I am hazarding a guess, I believe this should do the job A = { {1, x, 0}, {0, -1, 0}, {0, 0y, 1} } It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols B = { {0, 0, -\iota}, {\iota, 0, 0}, {0, 1, 0} } Then A.A= B.B.B = Id we can even choose x appropriately if needed the order of A.B should be infinite as the coefficient of the matrices are unbounded ? |
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