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Question 1201590: prove that, for square matrices A and B, AB=BA if and only if(A-B)(A+B)=A²-B².
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
The template P if and only if Q breaks into two pieces
For this problem,
AB = BA if and only if (A-B)(A+B) = A^2-B^2
breaks into- If AB = BA, then (A-B)(A+B) = A^2-B^2
- If (A-B)(A+B) = A^2-B^2, then AB = BA
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Part 1)
If AB = BA, then (A-B)(A+B) = A^2-B^2
We'll start with (A-B)(A+B) and try to reach A^2-B^2 through use of AB = BA
(A-B)(A+B) = A(A+B)-B(A+B)
(A-B)(A+B) = (A^2+AB)+(-BA-B^2)
(A-B)(A+B) = (A^2+AB)+(-AB-B^2) ... use AB = BA
(A-B)(A+B) = A^2+(AB-AB)-B^2
(A-B)(A+B) = A^2+0*AB-B^2
(A-B)(A+B) = A^2+0-B^2
(A-B)(A+B) = A^2-B^2
We have proven that if AB=BA, then (A-B)(A+B) leads to A^2-B^2
In other words, if AB=BA, then (A-B)(A+B) = A^2-B^2
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Part 2)
If (A-B)(A+B) = A^2-B^2, then AB = BA
(A-B)(A+B) = A^2-B^2
A(A+B)-B(A+B) = A^2-B^2
(A^2+AB)+(-BA-B^2) = A^2-B^2
A^2+(AB-BA)-B^2 = A^2-B^2
(AB-BA)-B^2 = -B^2
AB-BA = 0
AB = BA
The second portion of the "if and only if" statement has been confirmed.
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Therefore, overall we can say AB = BA if and only if (A-B)(A+B) = A^2-B^2
Matrices A and B must be square, and the same size.
Answer by ikleyn(52771)  (Show Source):
You can put this solution on YOUR website! .
(A-B)*(A+B) = A*A - B*A + A*B - B*B = (A^2 - B^2) + (A*B - B*A).
From this identity, which is valid for all matrices ALWAYS, you can easily conclude that
(A-B)*(A+B) = A^2 - B^2 if and only if A*B - B*A = 0.
It is the same as to say that
(A-B)*(A+B) = A^2 - B^2 if and only if A*B = B*A.
Solved.
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