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

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SUMMARY

An operator A that satisfies the equation A² - A + I = 0 is guaranteed to have an inverse. The inverse A⁻¹ can be expressed as a polynomial function of A. The discussion highlights the importance of the determinant being non-zero for the existence of an inverse and suggests that the secular equation may be a useful tool for solving related problems in quantum mechanics.

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  • Understanding of linear operators and their properties
  • Familiarity with determinants and their role in matrix inverses
  • Basic knowledge of polynomial functions of matrices
  • Introduction to quantum mechanics concepts, particularly secular equations
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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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Welcome to PF!

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.
 

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