# Let A and B are $2 \times 2$ matrices, where AB = BA, show that $\text{A B}^2 = \text{B}^2 \text{A}$

Let A and B are $2\times 2$ matrices, where $$\text{AB} = \text{BA}$$show that : $$\text{A B}^2 = \text{B}^2 \text{A}$$

Well, I assumed A and B are symmetric matrices, so $$AB^2 = A^T B pow(2T) = (B^ 2A)^T = B^2T = A^T = B^2 A$$ I thought for long time but I didn't get the right solution, any expert can help please!

## 4 Answers

I think the answer is a bit simpler than you expect. Suppose that $A,B$ are commuting matrices, that is, $AB = BA$. Then let's compute \begin{equation} AB^{2} = ABB = BAB = BBA = B^{2}A \end{equation} as desired.

Let's do it with making every step and used law explicit: \begin{align} AB^2 &= A(BB) && \text{definition of $B^2$}\\ &= (AB)B && \text{associativity}\\ &= (BA)B && \text{by assumption}\\ &= B(AB) && \text{associativity}\\ &= B(BA) && \text{by assumption}\\ &= (BB)A && \text{associativity}\\ &= B^2A && \text{definition of $B^2$} \end{align} Note that I didn't use anything specific to matrices; the same calculation works with any semigroup.

No need for symmetry.

Note that $AB = BA$ implies that $$AB^{2} = BAB$$ by right-multiplying $AB$ and $BA$ by $B$ and that $$BAB = B^{2}A$$ by left-multiplying $AB$ and $BA$ by $B$.

Take your equation $AB=BA$ and multiply both sides on the right by $B$. The left hand side becomes:

$$(AB)B = A(BB) = AB^2$$

and the right hand side becomes:

$$(BA)B = B(AB) = B(BA) = (BB)A = B^2A.$$

Thus, $$AB^2 = B^2A.$$