Prove $BA-A^2B^2=I$

Question

If $A$ and $B$ be two square matrices containing real elements and satisfying the following conditions

$$AB-B^2A^2=I$$

$$A^3+B^3=O$$

then prove that

$$BA-A^2B^2=I$$

My Attempt: If we are able to prove that if either of $A$ or $B$ is invertible then we are done.

score 1 · Accepted Answer · answered Sep 28 '17 at 14:57

1

Hint. What is $$ \pmatrix{A&-B^2\\ B^2&A}\pmatrix{B&-A^2\\ A^2&B}? $$

answered Sep 28 '17 at 14:57

user1551

It will be $$(AB-B^2A^2)I=I^2=I$$. Brilliant. – Maverick Sep 28 '17 at 15:38
I got it. Thanks a lot – Maverick Sep 28 '17 at 15:40
Motivated by your thinking , I developed the line $$(A-iB^2)(B-iA^2)=I$$ – Maverick Sep 28 '17 at 16:02
@Maverick Yes, that will do the trick too. – user1551 Sep 28 '17 at 16:06
Can you suggest any good book/resource from where more of such problems can be obtained – Maverick Sep 29 '17 at 05:00
@Maverick Sorry, I don't know any relevant resource. – user1551 Sep 29 '17 at 10:49

1 Answers1