0
$\begingroup$

If $A$ and $B$ are $n x n$ invertible matrices and $(AB)^2$ = $A^2B^2$, then $AB = BA$.

How would you go about proving or disproving this statement. I fail to connect my theorems to the $(AB)^2$ part, I can't think of a relation or hint.

$\endgroup$
  • 3
    $\begingroup$ you have $ABAB = A^2B^2$, simplify this expression. $\endgroup$ – Thoth Nov 1 '15 at 21:28
2
$\begingroup$

$(AB)^2=A^2B^2$ is equivalent to $ABAB=A^2B^2$. Multiplying to the left both members in the last expression by $A^{-1}$ (wich exists because A,B are invertible) you get $$BAB=AB^2$$ then multiplying to the right both members by $B^{-1}$ you get $$BA=AB.$$

$\endgroup$
2
$\begingroup$

If $A,B$ are invertible we have: $$ (AB)^2=A^2B^2 \quad \iff \quad ABAB=AABB \quad \iff \quad A^{-1}ABABB^{-1}=A^{-1}AABBB^{-1} \quad \iff \quad BA=AB $$

$\endgroup$

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.