31
$\begingroup$

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

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

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

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

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

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

$$B^{16}=BA^4=B$$

is there any other approach to solve this?

$\endgroup$
2
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – davidlowryduda
    Oct 8, 2017 at 3:31
  • 5
    $\begingroup$ 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, 2017 at 0:28

5 Answers 5

101
+250
$\begingroup$

$A^4=I$ 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 $B^{16}=B$.

$\endgroup$
11
  • 5
    $\begingroup$ Indeed .. $(B^2)^2= (A^{-1}BA)(A^{-1}BA)=A^{-1}(B^2)A=(A^{-1})^2BA^2$ $\endgroup$
    – Widawensen
    Jun 22, 2017 at 11:21
  • 3
    $\begingroup$ For a complete proof and a generalization, see the egreg's post in math.stackexchange.com/questions/2277632/… $\endgroup$
    – user91684
    Jul 28, 2017 at 9:07
  • 1
    $\begingroup$ More precisely, $A^{-1}=A^3$. $\endgroup$ Sep 21, 2018 at 17:38
  • 5
    $\begingroup$ 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. $\endgroup$
    – user459879
    Aug 15, 2019 at 12:30
  • 1
    $\begingroup$ @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? $\endgroup$ Aug 16, 2019 at 16:51
32
$\begingroup$

$$B^2=A^4B^2=A^3BA.$$ 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 $$B^{16}=ABA^3ABA^3=AB^2A^3=BA^4=B$$

$\endgroup$
4
  • $\begingroup$ awesome sir thanks good approach $\endgroup$ Jun 22, 2017 at 10:23
  • $\begingroup$ @Ekaveera Kumar Sharma Welcome! $\endgroup$ Jun 22, 2017 at 10:27
  • $\begingroup$ Perhaps the first line should be $B^2=IB^2=A^4B^2 ...$ to be more explicit? $\endgroup$ Sep 21, 2018 at 17:40
  • $\begingroup$ @Acccumulation Yes, of course, but it's obvious I think. $\endgroup$ Sep 21, 2018 at 17:43
30
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ I think the starting question is a typical abstract-algebraic question. What do you think about it? Thank you! $\endgroup$ Oct 11, 2017 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, 2017 at 8:52
  • $\begingroup$ Thank you! Just there are people, which think another. $\endgroup$ Oct 16, 2017 at 8:59
6
$\begingroup$

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 $3,7$.

$\endgroup$
5
$\begingroup$

My Method:

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

$$A^4B^2=A^3BA$$

$$B^2=A^3BA \tag{1}$$ 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

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

$\endgroup$
9
  • $\begingroup$ How do you get $(2)$ from $(1)$? $\endgroup$
    – Soham
    Sep 21, 2018 at 17:42
  • $\begingroup$ timessimply by multipling it $\endgroup$ Sep 21, 2018 at 17:44
  • $\begingroup$ Multiplying what by what? $\endgroup$
    – Soham
    Sep 21, 2018 at 17:46
  • $\begingroup$ let me show u the problem $\endgroup$ Sep 21, 2018 at 17:46
  • $\begingroup$ $B^2*B^2*B^2*B^2*B^2*B^2*B^2*B^2$=...... $\endgroup$ Sep 21, 2018 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.