If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

Question

Matrix Question

Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!

Thanks for help in advance!

score 13 · Accepted Answer · answered May 11 '13 at 12:26

13

$$ \begin{align} \color{#C00000}{BA} &=AA(\color{#C00000}{BA})BB\\ &=\color{#00A000}{A}(ABAB)\color{#00A000}{B}\\ &=\color{#00A000}{AB} \end{align} $$

answered May 11 '13 at 12:26

robjohn

345,667

score 2 · Answer 2 · answered May 11 '13 at 12:06

2

Multiplying $A(BA-AB)B$ on the left by $A$ and the right by $B$ gives $$AB-BA=A^2(BA-AB)B^2=A0B=0$$ as desired.

answered May 11 '13 at 12:06

Alex Becker

60,569

score 1 · Answer 3 · answered May 11 '13 at 12:06

1

We know $AA = I, BB= I, ABAB = I$. Thus $A^{-1} = A, B^{-1} = B$. Now $$ABAB = I \\ AB = IB^{-1}A^{1} = BA$$

answered May 11 '13 at 12:06

Stefan

3,451

If $A^2 = I$, then $A$ is invertible and its inverse is $A$. – Stefan May 11 '13 at 12:18
From above: have not learnt inverses yet! Guess that's why I didn't think of this... – user73229 May 11 '13 at 12:18

score 1 · Answer 4 · answered May 11 '13 at 12:06

1

Hint: Is $BA$ the inverse of $AB$. How to check that?

answered May 11 '13 at 12:06

Bunder

2,423

Have not learnt inverses yet! – user73229 May 11 '13 at 12:18

score 0 · Answer 5 · answered May 11 '13 at 12:19

0

$$BA=BAI=BA(ABAB)=B(AA)BAB=BIBAB=(BB)AB=IAB=AB$$

answered May 11 '13 at 12:19

Nikolaj-K

12,249

score 0 · Answer 6 · answered May 24 '13 at 05:47

0

$A,B$ belong to the group $GL_n(F),~(F$ being the field over which $A,B$ are defined$).$ So you may apply cancellation on $(AB)^2=A^2B^2.$

answered May 24 '13 at 05:47

Sugata Adhya

3,979

If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$

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