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

[Kavorka]

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.

 

[jfizzix]

If AB = BA
and (A+B)^2 = (A+B)(A+B)
then the rest more or less falls into place.

 

[Kavorka]

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

 

[Mark44]

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?

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).

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

 

[WWGD]

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.