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

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^-1A.AA^-1A...

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 :s-smilie:

So for A and B to be symmetric, does this mean A=B?
And for (b) do I have to prove both ways

Reply 1

ghostwalker

17

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^-1A.AA^-1A...

So you're looking at

A^3

and

A+I

. 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.

(edited 10 years ago)

Reply 2

#maths

OP

Original post by ghostwalker

So you're looking at

A^3

and

A+I

. 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.