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# show (I+A) is invertible and AB is symmetric watch

1. Let A and B be square matrices of the same size. Show the following:

(a) If A3 = 0 then (I + A) is invertible and (I + A)-1 = I - A + A2.

for this I started off: A3 = AAA = IA.IA.IA = AA-1A.AA-1​A.AA-1​A...

then I got stuck

(b) If A and B are symmetric then AB is symmetric if and only if A and B commute.

I have no idea where to start for this one
So for A and B to be symmetric, does this mean A=B?
And for (b) do I have to prove both ways
2. (Original post by #maths)
Let A and B be square matrices of the same size. Show the following:

(a) If A3 = 0 then (I + A) is invertible and (I + A)-1 = I - A + A2.

for this I started off: A3 = AAA = IA.IA.IA = AA-1A.AA-1​A.AA-1​A...
So you're looking at and . Does this not suggest doing something with the sum of two cubes, particularly given the inverse.

Also you may want to start using Latex if you're going to be posting frequently. There's a link at the top of the forum.
3. (Original post by ghostwalker)
So you're looking at and . Does this not suggest doing something with the sum of two cubes, particularly given the inverse.

Also you may want to start using Latex if you're going to be posting frequently. There's a link at the top of the forum.
Could you elaborate?
4. (Original post by #maths)
could you elaborate?

Now factorise the RHS.
5. It is enough to show that (I+A)(I-A+A^2)=(I-A+A^2)(I+A)=I
6. (Original post by james22)
It is enough to show that (I+A)(I-A+A^2)=(I-A+A^2)(I+A)=I
So true.

I made a slip multipliying the two, which is why I went for the factorisation. Ho, hum.

@OP: As james 22 said.
7. (Original post by james22)
It is enough to show that (I+A)(I-A+A^2)=(I-A+A^2)(I+A)=I
(Original post by ghostwalker)
So true.

I made a slip multipliying the two, which is why I went for the factorisation. Ho, hum.

@OP: As james 22 said.
Thanks for your help. I've been making everything seem more complicated in my head :/

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