Can (I+A)^-1 be simplified to I+A/2 in linear algebra?

[isa_vita]

A is a square matrix n*n with the following properties: A*A=A and A not equal I (identity matrix).
How to prove the following equation:
(I+A)^-1=I+A/2 ?

 

[HallsofIvy]

You can't. It's not true:
(I+ A)(I+ A/2)= I+ A+ A/2+ A^2/2= I+ A+ A/2+ A/2= I+ 2A, not I.

If you mean (I+ A)/2,
(I+ A)(I+ A)/2= (I+ A+ A+ A^2)/2= (I+ 3A)/2.

I thought perhaps you had left out a "-" but that doesn't see to work either so I have no idea what you are trying to prove.

 

[isa_vita]

I mean with this (I+A)^-1, inverse of the matrix (I + A).
You have understood correctly.
As I tried to solve the problem I came to the same answer.
Perhaps there is a technical error, but I'm not sure.
Thanks for answer.

 

[oli4]

Hi isa_vita, maybe this is coming from language issues, or maybe there is an error in the exercise you want to do, could you post the text as it comes ? (translated) ?
you equation does have a solution: A=0 works, but you would say 'solve' instead of 'prove', so maybe you didn't word it correctly (but it sounds fishy, this would be a weird exercise anyway), so, anyway, do you have the full text of the exercise ?
Cheers...

 

[isa_vita]

Hi oli4e, my mistake, I mean prove not solve. I gave full text of the exercise(in my translation .
I know for a zero solution, but it isn't the task. I think it is right HallsofIvy. It is a technical error.
Thanks for a help. Cheers...

 

[HallsofIvy]

I still don't understand what you are asking. Given that $A^2= A$, it does NOT follow that I+ A/2 is the inverse of I+ A.

 

[chiro]

Assuming that the inverse even exists why not just multiply both sides by (I + A) and collect terms? (Remember I'm assuming that the inverse of this exists which it may not, since I haven't checked it).

 

[HallsofIvy]

Assuming that the inverse even exists why not just multiply both sides by (I + A) and collect terms? (Remember I'm assuming that the inverse of this exists which it may not, since I haven't checked it).

That's exactly what I did- and did NOT get the identity matrix.