# M^2 = I

1. Apr 5, 2007

### Ultraworld

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.

Last edited: Apr 5, 2007
2. Apr 5, 2007

### matt grime

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.

3. Apr 5, 2007

### Ultraworld

Last edited: Apr 5, 2007
4. Apr 5, 2007

### Ultraworld

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

Last edited: Apr 5, 2007
5. Apr 5, 2007

### matt grime

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.

6. Apr 5, 2007

### Ultraworld

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

7. Apr 5, 2007

### Ultraworld

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?

Last edited: Apr 5, 2007
8. Apr 5, 2007

### matt grime

Yes. Obviously.

9. Apr 5, 2007

### HallsofIvy

Staff Emeritus
Any determinant has at least one eigenvalue (in the complex numbers) because every polynomial has at least one complex solution. Since the determinant is not 0, that eigenvalue is not 0, therefore ---