Can a Linear Operator Satisfying A^2 - A + I = 0 Always Have an Inverse?

seek
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Homework Statement



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.

Homework 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

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.
 
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Hi seek! Welcome to PF! :smile:
seek said:
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.

oooh … so complicated:cry:

Try writing it A2 - A = -I :wink:
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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