# Deducing M^2 = I

Given an n by n matrix M with complex coefficients such that M2 = I and M is not equal to I.

What can I deduce from it. e.g. what does it say about the eigenvalues?

edit: of course M -1 = M.

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matt grime
Homework Helper
M satisfies the polynomial x^2-1. M does not satisfy x-1. That tells you that the minimal poly is either x^2-1 or x+1. That tells you all of the eigenvalues.

So -1 is an eigenvalue of M.

However am I guaranteed that the equation M x = - x has a non-trivial solution?

EDIT: I think the answer is yes cause the dimension of the Eigenspace is the same as the geometric multiplicity of -1 http://en.wikipedia.org/wiki/Eigenvalue (see definitions)

EDIT2: Im sure the answer is yes

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matt grime
Homework Helper
If -1 is not an eigen value of M, then all its eigenvalues are 1, and it must satisfy x-1, so M would be the identity matrix, which we are told it is explicitly not.

By definition, if t is an eigenvalue of M, then there is an eigenvector with eigenvalue t.

thanks Matt. This was part of a bigger problem and that bigger one is now solved.

I can make my solution much easier.

Given a n by n matrix N with complex coefficients and det N != 0. Does N has a non-zero eigenvalue?

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matt grime