Prove or show the statement is false with counterexample:
Square matrices $A$ and $B$ commute if and only if $(A-B)(A+B)=A^2-B^2$

Now, this is a biconditional statement, so we must prove $X=>Y$ $\land$ $Y=>X$

I managed to prove $X=>Y$: if the statement $(A-B)(A+B)=A^2-B^2$ holds, the two matices must commute but how do you prove, or disprove $Y=>X$: if matrices commute the statement $(A-B)(A+B)=A^2-B^2$ holds?
Here is the part of the proof:

$(A-B)_{ik}(A+B)_{kj}$ =$\sum\limits_{k=1}^\mathbb{n}(A-B)_{ik}(A+B)_{kj}$




From this we can see if the statement $(A-B)(A+B)=A^2-B^2$ holds, $AB_{ij}-BA_{ij}=0$ which is true only if matrices $A$ and $B$ commute.

How would you now prove or disproove that if matrices $A$ and $B$ commute $(A-B)(A+B)=A^2-B^2$ must be the case?


You're thinking way too overcomplicated. We have $$(A+B)(A-B)=A^2-AB+BA-B^2.$$ Both directions follow at once.

  • $\begingroup$ I just realised the statement in the title is not the same as the one in the body. Can the OP clarify which one is meant? $\endgroup$ – YiFan Oct 6 at 10:22
  • $\begingroup$ What part is not the same? $\endgroup$ – ToTheSpace 2 Oct 6 at 10:28
  • $\begingroup$ @ToTheSpace2 Are we supposed to show $AB=BA$ if and only if $(A+B)(A-B)=A^2-B^2$ or $(A+B)(A-B)=A^2+B^2$? $\endgroup$ – YiFan Oct 6 at 10:29
  • $\begingroup$ My mistake… I didn't notice the mistake. I corrected the mistake. $\endgroup$ – ToTheSpace 2 Oct 6 at 10:32
  • 1
    $\begingroup$ Thank you very much for your help! I think I understand everything with regards to this proof now. Thank you once more. $\endgroup$ – ToTheSpace 2 Oct 6 at 10:40

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