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Let $A, B$ be two invertible matrices of order $r$ such that

$$ABA=BA^2B, \qquad A^3=I, \qquad B^{2n-1}=I \text{ for some positive integer } n$$

I am interested in checking if $A$ and $B$ are commutative. Also if $B$ is idempotent ($B^2=B$) or involutory ($B^2=I$).


My approach

As $$A^3=I, A^2=A^{-1}$$ $$ABA=BA^2B=BA^{-1}B$$ $$A^{-1}B^{-1}A^{-1}=B^{-1}AB^{-1}$$ Now I am unable to proceed from here. I thought $$B^{-1}A^{-1}=AB^{-1}AB^{-1}$$ But this also doesn't yield any fruitful result. If anyone can share some alternate ways to these type of problems for faster solving or if anyone can spot how to proceed with this, it would be a great help. Thank You.

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  • $\begingroup$ So you want to compute the element $[A,B]$ and the order of $B$ in the group $\langle A,B\mid ABAB^{-1}AB^{-1}, A^3, B^{2n-1}\rangle$ which is just an example of word problem. $\endgroup$ Aug 26 at 8:12
  • $\begingroup$ @user10354138, I haven't studied this (word problem) yet. $\endgroup$
    – PCMSE
    Aug 26 at 8:45
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This is essentially a word problem: $\left \langle A, B \mid ABA = BA^2B, A^3 = I, B^{2n-1} = I \right \rangle$. A method I commonly use is to utilise properties of similar matrices. We have \begin{align*} ABA &= BA^2B\\ B^{-1}AB &= A^2BA^{-1}\\ (B^{-1}AB)^3 &= (A^2BA^{-1})^3\\ (B^{-1}AB)(B^{-1}AB)(B^{-1}AB) &= (A^2BA^{-1})(A^2BA^{-1})(A^2BA^{-1})\\ B^{-1}A^3B &= A^2BABABA^{-1}\\ I &= A^2BABABA^{-1}\\ I &= ABABAB\\ &= (ABA)(BAB)\\ &= BA^2B^2AB\\ B^{-2} &= A^{-1}B^2A\\ B^{-2n} &= (A^{-1}B^2A)^n\\ B^{-1} &= A^{-1}B^{2n}A\\ &= A^{-1}BA\\ A^2 &= BA^2B\\ &= ABA\\ I &= B \end{align*} Hence, we must have $B = I$, and $A$ is any matrix of order $3$. This does give that $A$ and $B$ commute, with $B$ both idempotent and involutory.

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  • $\begingroup$ Can you please elaborate the 4th step after taking cube? $\endgroup$
    – PCMSE
    Aug 26 at 9:14
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    $\begingroup$ @PCMSE I've expanded it out for you, hope that helps. Essentially, by cubing, I'm getting terms to cancel out with their inverse - you might've seen this when calculation powers of matrices once you know their Jordan Normal Form. $\endgroup$ Aug 26 at 9:17
  • $\begingroup$ Thank you very much @SharkyKesa :) Understood clearly $\endgroup$
    – PCMSE
    Aug 26 at 9:18
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I would have started by looking at what happens if we assume one of the properties you want to show. For example $AB = BA$ together with $ABA = B A^2 B$ clearly implies $A^2 B = A^2 B^2$ and thus $B = I$. Similarly, if you assume $B^2 = I$, then $B^{2n - 1} = I$ implies $B = B^{2n} = (B^2)^n = I$. Finally note that $B^2 = B$ implies $B = I$ without any additional assumptions since $B$ was assumed to be invertible.

So now we know that $A$ and $B$ only commute if the relations given already imply that $B = I$ and of course if we have $B = I$, then all the desired relations ($AB = BA$, $B^2 = B$ and $B^2 = I$) are satisfied as well.

Now, we can look for an example of matrices $A$ and $B$ which satisfy the above relations as well as $B \neq I$ and then we would know that the relations do not imply any of the relations mentioned.

At the same time, we can try to do what you already started and look for new relations that might help us maybe deduce $B = I$ from the given relations. In fact, we have $$A B A^{-1} = A B A^2 = B A^2 B A = BA (ABA) = B A B A^2 B = B (ABA^{-1}) B$$ And so we have $(ABA^{-1}) B (ABA^{-1})^{-1} = B^{-1}$ and this generalizes to $$(ABA^{-1})^k B (ABA^{-1})^{-k} = B^{(-1)^{k}}$$ for all $k$. However, for $k = 2n-1$ we have $(ABA^{-1})^k = A B^{k} A^{-1} = I$ and so $B = B^{(-1)^{2n-1}} = B^{-1}$ which gives $B^2 = I$ and as we saw earlier, using $B^{2n-1} = I$ this implies $B = I$.

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