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

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?

Solution 1:

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

Solution 2:

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

Solution 3:

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.

Solution 4:

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 $3,7$.