It is not right. Is is true if is hemitian i.e. or more general, it is true if is normal i.e.
Regards.
http://www.fernandorevilla.es/
Find all symmetric matrices A such that A^2011 = I. I diagonalized A so that A = CDC^-1 where D is a diagonal matrix composed of the eigenvalues of A. since A is symmetric, C^-1 = C^T so C is an orthogonal matrix. so i know that a^2011 = C D^2011 C^-1 = I or that D^2011 = I. this means that each of the lambda's in D have an equation of form λ^2011 = 1. since it was not given that it was a real matrix, i cannot assume that 1 will be the one root so i also include the complex roots. these are cos(2πk/2011) + i sin(2πk/2011) where k = 0, 1, ...2010. so to find all symmetric matrices A i said that you take a diagonal matrix whose entries on the diagonal are any of the 2011(th) roots of unity and multiply it on the left by C^T and on the right by C where C is any orthogonal matrix.
is this correct? is it right to assume that A is diagonalizable even though it is not necessarily real (as in this problem)? i know that the spectral theorem guarantees diagonalizability for real and symmetric matrices and guarantees that all the eigenvalues are real. if the matrix is symmetric but not composed solely of real entries, can it still be diagonalizable?
It is not right. Is is true if is hemitian i.e. or more general, it is true if is normal i.e.
Regards.
http://www.fernandorevilla.es/
thank you for your feedback.
Find all symmetric matrices A such that A^2011 = I. (Diagonalize and find the eigenvalues of A). Sorry, i forgot the last part of the problem which told me to diagonalize so that i why i did it. but if that last part in the parenthesis was not there, i cannot just assume A is diagonalizable and do the problem as i posted right?
also, was my method of doing the problem correct? would any A such that A^2011 = I be matrices of the form C^T D C where C is any othogonal matrix and D is a diagonal matrix containing any combination of the 2011(th) roots of unity?
In that case I am pretty sure that and we needn't the Spectral Theorem:
Then,
Regards.
Fernando Revilla