If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.

My Method:

Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get

$$AB^4=BAB^2=B^2A$$ Hence

$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we get

$$A^2B^4=(AB^2)A=BA^2$$ hence

$$A^2B^4=BA^2 \tag{2}$$ Now post multiplying with $B^4$ and using $(2)$we get

$$A^2B^8=B(A^2B^4)=B^2A^2$$ hence

$$A^2B^8=B^2A^2 \tag{3}$$

Now Pre Multiply with $B^2$ and use $(3)$ we get

$$B^2A^2B^8=B^4A^2$$ $\implies$



Now pre multiply with $A^2$ and use $(2)$we get

$$A^4B^{16}=A^2B^4A^2$$ $\implies$


is there any other approach to solve this?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – davidlowryduda Oct 8 '17 at 3:31
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    $\begingroup$ While it's good to tag a question appropriately, please refrain from getting into an edit war, especially a long one. Please be reminded that every time a question is re-tagged, it will be bumped up to the front page. $\endgroup$ – user1551 Oct 9 '17 at 0:28

$$B^2=A^4B^2=A^3BA.$$ Thus, $$B^4=A^3BAA^3BA=A^3B^2A=A^2BA^2.$$ Hence, $$B^8=A^2BA^2A^2BA^2=A^2B^2A^2=ABA^3$$ and from here $$B^{16}=ABA^3ABA^3=AB^2A^3=BA^4=B$$

  • $\begingroup$ awesome sir thanks good approach $\endgroup$ – Ekaveera Kumar Sharma Jun 22 '17 at 10:23
  • $\begingroup$ @Ekaveera Kumar Sharma Welcome! $\endgroup$ – Michael Rozenberg Jun 22 '17 at 10:27
  • $\begingroup$ Perhaps the first line should be $B^2=IB^2=A^4B^2 ...$ to be more explicit? $\endgroup$ – Acccumulation Sep 21 '18 at 17:40
  • $\begingroup$ @Acccumulation Yes, of course, but it's obvious I think. $\endgroup$ – Michael Rozenberg Sep 21 '18 at 17:43

$A^4=I$ implies that $A$ is invertible. Hence $B^2=A^{-1}BA$ . Repeatedly squaring and using the previous step we get $B^{16}=A^{-4}BA^{4}$ which gives $B^{16}=B$.

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    $\begingroup$ Indeed .. $(B^2)^2= (A^{-1}BA)(A^{-1}BA)=A^{-1}(B^2)A=(A^{-1})^2BA^2$ $\endgroup$ – Widawensen Jun 22 '17 at 11:21
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    $\begingroup$ For a complete proof and a generalization, see the egreg's post in math.stackexchange.com/questions/2277632/… $\endgroup$ – loup blanc Jul 28 '17 at 9:07
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    $\begingroup$ More precisely, $A^{-1}=A^3$. $\endgroup$ – Acccumulation Sep 21 '18 at 17:38
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    $\begingroup$ Funny how people feel the need to argue semantics rather than maths. We iteratively square and substitute in some boot-strapping manner, inductively until we arrive at the desired result. $\endgroup$ – Wesley Strik Aug 15 '19 at 12:30
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    $\begingroup$ @WesleyStrik Kavi asked why someone downvoted (I guess he deleted that comment) and I gave my guess. Another benefit of my comments is that they increase the probability kavi edits his answer, making the math more clear for somebody else. Does this make sense? $\endgroup$ – mathworker21 Aug 16 '19 at 16:51

The relation $AB^2 = BA$ allows to reduce the number of $B$s by one. Using this idea, we can show that for every even power $2n$, we have that $$AB^{2n} = B^nA.$$

Applying this and using $A^4 = I$, we can simplify $$B^{16} = A^4B^{16} = A^3B^8A = \ldots = B.$$

When we pull the first $A$ through, we half the power of $B$. Now it should be easy to see how to continue and why we will be left with $B$ in the end.

Note that you used basically the same idea, you just hid it behind multiple steps.

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    $\begingroup$ I think the starting question is a typical abstract-algebraic question. What do you think about it? Thank you! $\endgroup$ – Michael Rozenberg Oct 11 '17 at 14:13
  • $\begingroup$ @MichaelRozenberg Not sure why you're asking, but yes, it is a introductory level algebra question. You have some properties and have to apply them in a clever way to get a desired result; and you might use a little induction. I see the question as introductory level (both to algebra and proofs in general) as you only need to manipulate the given information. You don't need any more knowledge, you don't even need $A$ and $B$ to be matrices, it's just a very basic proof to get a desired information from predetermined facts. Why do you ask? :) $\endgroup$ – Dirk Oct 16 '17 at 8:52
  • $\begingroup$ Thank you! Just there are people, which think another. $\endgroup$ – Michael Rozenberg Oct 16 '17 at 8:59

Let $p\geq 2$. The interesting question is: what is $k_n$ where

$k_n=\min\{k\geq 2|A,B\in M_n(\mathbb{C}),AB^2=BA,A^p=I_n\Rightarrow B^k=B\}$ ? According to the other answers, $B^{2^p}=B$; then $k_n\leq 2^p$.

Step 1. $k_n$ is reached for some root of unity $B$.

Proof. $B^2$ and $B$ is similar; then $\ker(B^2)=\ker(B)$ and $0$ is a semi simple eigenvalue of $B$ (or $B$ is invertible). We may assume that $B=diag(U_q,0_r),A=\begin{pmatrix}P_q&Q\\R&S_r\end{pmatrix}$ where $U$ is invertible. We obtain $PU^2=UP,RU^2=0,UQ=0$; then $R=0,Q=0$ and $A=diag(P,S)$ where $P^p=I,S^p=I$. Therefore the problem reduces to the case when $B=U$ is invertible, as will be assumed in the sequel. Then $k_n=\min\{k\geq 2|A,B\in GL_n(\mathbb{C}),AB^2=BA,A^p=I_n\Rightarrow B^k=B\}$. Since $B$ is invertible, $B^{2^p-1}=I$ and $B$ is a root of unity. $\square$

Step 2. Let $\sigma(B)$ be the spectrum of $B$ and $\lambda\in \sigma(B)$. Since $B,B^2$ are similar, $\lambda^2,\lambda^{2^2},\cdots\in\sigma(B)$; it is not difficult to deduce that there is $s\leq n$ s.t. $\lambda^{2^s-1}=1$. We seek a solution $A,B$ s.t. $order(B)$ (which is a divisor of $2^p-1$) is maximal.

Case 1. $n\geq p$. Then $k_n=2^p$.

Proof. Let $\omega$ be a primitive $(2^p-1)^{th}$-root of unity. Take $B=diag(\omega,\omega^2,\cdots,\omega^{2^{p-1}},I_{n-p})$ and $A=diag(V,I_{n-p})$; here $V=[v_{i,j}]$ is the permutation defined by: the $v_{i,j}$ are $0$ except $v_{i,i+1}=1,v_{n,1}=1$.

Case 2. $n<p$. Then $k_n$ may be $<2^p$.

Proof. For instance, let $p=6,n=4$. $Order(B)$ is a divisor of $63$. The orders of the eigenvalues of $B$ are divisors of the $2^s-1,s\leq 4$, that are $1,3,7,15$. We cannot do better than $order(B)=7$ (therefore $k_4=8$) with $B=diag(\omega,\omega^2,\omega^4,1)$ where $\omega$ is a primitive $7^{th}$-root of unity.

Let $p=6,n=5$. In the same way, the orders of the eigenvalues of $B$ are divisors of $1,3,7,15,31$ and we obtain $k_4=22$ with $B=diag(u,u^2,v,v^2,v^4)$ where $u,v$ are primitive roots of orders $3,7$.


My Method:

Given $$AB^2=BA $$ Pre multiplying with $A^3$ we get


$$B^2=A^3BA \tag{1}$$ Using $(1)$ we get

$$B^{16}=A^3B^8A \tag{2}$$ $$B^{8}=A^3B^4A \tag{3}$$ $$B^{4}=A^3B^2A \tag{4}$$ $$B^{2}=A^3BA \tag{5}$$

Now combining $(5)$ in $(4)$ , $(4)$ in $(3)$ , $(3)$ in $(2)$ and we get

$$B^{16}=A^{12}BA^4$$ $$B^{16}=IBI$$ $$B^{16}=B$$

  • $\begingroup$ How do you get $(2)$ from $(1)$? $\endgroup$ – tatan Sep 21 '18 at 17:42
  • $\begingroup$ timessimply by multipling it $\endgroup$ – Marvel Maharrnab Sep 21 '18 at 17:44
  • $\begingroup$ Multiplying what by what? $\endgroup$ – tatan Sep 21 '18 at 17:46
  • $\begingroup$ let me show u the problem $\endgroup$ – Marvel Maharrnab Sep 21 '18 at 17:46
  • $\begingroup$ $B^2*B^2*B^2*B^2*B^2*B^2*B^2*B^2$=...... $\endgroup$ – Marvel Maharrnab Sep 21 '18 at 17:49

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