# How to prove or disprove matrices $A$ and $B$ commute iff $(A-B)(A+B)=A^2-B^2$?

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:
$$(A-B)_{ik}=a_{ik}b_{jk}$$
$$(A+B)_{kj}=a_{kj}b_{kj}$$

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

=$$\sum\limits_{k=1}^\mathbb{n}(a_{ik}-b_{ik})(a_{kj}+b_{kj})$$

=$$\sum\limits_{k=1}^\mathbb{n}a_{ik}a_{kj}+a_{ik}b_{kj}-n_{ik}a_{kj}-b_{ik}b_{kj}$$

=$$\sum\limits_{k=1}^\mathbb{n}a_{ik}a_{kj}+\sum\limits_{k=1}^\mathbb{n}a_{ik}b_{kj}+\sum\limits_{k=1}^\mathbb{n}-b_{ik}a_{kj}+\sum\limits_{k=1}^\mathbb{n}-b_{ik}b_{kj}$$

=$$A^2_{ij}-B^2_{ij}+AB_{ij}-BA_{ij}$$
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.
• @ToTheSpace2 Are we supposed to show $AB=BA$ if and only if $(A+B)(A-B)=A^2-B^2$ or $(A+B)(A-B)=A^2+B^2$? – YiFan Oct 6 at 10:29