Proof of (A+B)^2 = A^2 + 2AB + B^2 for Matrix Algebra

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Discussion Overview

The discussion revolves around proving the identity (A+B)^2 = A^2 + 2AB + B^2 for n x n matrices A and B under the condition that AB = BA. Participants explore the manipulation of matrix expressions and the implications of the distributive property in matrix algebra.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Some participants suggest that the proof is straightforward, relying on the distributive property and the commutativity of A and B.
  • Others emphasize the importance of the condition AB = BA in the proof, questioning whether it is necessary to state this explicitly.
  • Several participants propose that expanding (A+B)(A+B) leads directly to the desired identity.
  • There are mentions of the identity matrix I and its properties, particularly in the context of the expression (I+A)(I+A) and its expansion.
  • Some participants express uncertainty about the level of detail required in the proof and whether their explanations meet the expectations of the problem.

Areas of Agreement / Disagreement

Participants generally agree on the basic steps to prove the identity, but there is some uncertainty regarding the necessity of detailing the commutativity condition and the level of detail required in the proof. No consensus is reached on how much elaboration is needed.

Contextual Notes

Some participants note that the proof relies on the assumption that AB = BA, which may not be explicitly stated in all steps. There is also a mention of the need for clarity in expanding expressions involving the identity matrix.

Who May Find This Useful

This discussion may be useful for students studying matrix algebra, particularly those working on proofs involving matrix identities and properties of commutative matrices.

Kavorka
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This problem is so simple that I'm not exactly sure what they want you to do:

Let A and B be n x n matrices such that AB = BA. Show that (A + B)^2 = A^2 + 2AB + B^2. Conclude that (I + A)^2 = I + 2A + A^2.

We don't need to list properties or anything, just manipulate. This all seems self-evident from the distributive property, and showing that I^2 = I.
 
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If AB = BA
and (A+B)^2 = (A+B)(A+B)
then the rest more or less falls into place.
 
So AB = BA
(A+B)(A+B) = A^2 + 2AB + B^2
AI=IA=A
II = I
(I+A)(I+A) = I^2 + 2AI + A^2 = I + 2A + A^2

Would this probably be what they're looking for? Not sure how much more in detail I can go
 
Kavorka said:
So AB = BA
(A+B)(A+B) = A^2 + 2AB + B^2
I think you need some more detail here. Is it important that AB = BA in your proof?
Kavorka said:
AI=IA=A
II = I
(I+A)(I+A) = I^2 + 2AI + A^2 = I + 2A + A^2
I think you need some more detail here as well, particularly in how you expand (I + A)(I + A).
Kavorka said:
Would this probably be what they're looking for? Not sure how much more in detail I can go
 
Kavorka said:
So AB = BA
(A+B)(A+B) = A^2 + 2AB + B^2
AI=IA=A
II = I
(I+A)(I+A) = I^2 + 2AI + A^2 = I + 2A + A^2

Would this probably be what they're looking for? Not sure how much more in detail I can go

First of all, great name. Then, like Mark said, just expand the product term-by-term, without grouping.
 

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