For square matrices $ (AB)^k=I_n$ for some $k$ implies $(BA)^k=I_n$

Question

Let $A$ and $B$ be $n \times n$ square real matrices.

Suppose $(AB)^k=I_n$ for some $k$ then $(BA)^k=I_n$

I know the result holds for $k=1$. I believe this should be easy but I cannot see it. Any ideas?

How would you show it for $k=1$? With a little luck, that proof can be generalized with only minor adjustments. — Arthur, Mar 12 '21 at 07:40
See https://math.stackexchange.com/q/238212/42969 – found quickly with Approach0 — Martin R, Mar 12 '21 at 07:50

score 1 · Accepted Answer · answered Mar 12 '21 at 07:43

1

First, try to prove that $B$ is invertible. To do this, suppose that $B$ is not invertible, i.e. $Bx = 0$ for some non-zero vector $x$, and try to arrive at a contradiction. There might be an easier way to do this too.

Then, take the equation $(AB)^k = I_n$, i.e. $$\underbrace{ABAB\cdots ABAB}_{k \ AB\text{'s}} = I_n$$ and multiply both sides by $B$ on the left and $B^{-1}$ on the right. What do you get?

answered Mar 12 '21 at 07:43

JimmyK4542

54,331

Thank you very much, it was easy! – Math Lover Mar 12 '21 at 07:52

For square matrices $ (AB)^k=I_n$ for some $k$ implies $(BA)^k=I_n$

1 Answers1