# Inverse of a linear operator

1. Feb 2, 2009

### seek

1. The problem statement, all variables and given/known data

Show that if an operator A satisfies A2 - A + I = 0 then A has an inverse. Express A-1 as a simple polynomial of A.

2. Relevant equations

I'm not sure that this is relevant, but A-1=1/(detA)TrC where TrC is the transpose of the matrix of cofactors. Also:
If detA = 0 then the matrix has no inverse

3. The attempt at a solution
So I notice immediately that adding by the identity matrix in this equation will result in a matrix with its diagonal having numbers (real or complex) and the rest being zero, as I can be expressed as the kronecker delta. And if the determinant must be nonzero in order to have an inverse, there has to be a way to relate the diagonal of an n dimensional matrix with its determinant. I'm just stuck as to how to do that. Any help greatly appreciated, thank you.
*Edit:
I've been thinking more about this problem, and it seems like there should be a way to use the secular equation to solve it. We went over it briefly in class (the class is quantum and I haven't had linear algebra yet, so it's kind of a chore), but not in enough detail that I would be able to use it in a proof.

Last edited: Feb 2, 2009
2. Feb 2, 2009

### tiny-tim

Welcome to PF!

Hi seek! Welcome to PF!
oooh … so complicated

Try writing it A2 - A = -I

3. Feb 3, 2009

### seek

My oversight is to my pride as a cold slap to the visage. Thanks so much for the help, maybe next time I'll be able to use the skills I learned in 5th grade.