Let T be a linear operator on the vector space of nxn matrices on the real field, defined by T(A) = transpose A.
Show that +/- 1 are the only eigenvalues of T, and describe corresponding eigenvectors.
The characteristic polynomial is given by f(t) = det ( [T]b - tIn) = 0, where [T]b is the matrix representation of T on some basis b and In is the nxn identity matrix. The eigenvalues should be given by the roots to f(t).
Also, there are only eigenvalues if det ([T]b - tIn) = 0.
The Attempt at a Solution
The only thing I can really see is that for any nxn matrix A, its transpose will have the same diagonal elements as it. But since I'm talking about the matrix representation of T, not matrix A, I'm not really sure how this helps. If we were to choose a basis and write [T]b, I am also fairly sure that this wouldn't be an upper triangular matrix, so I don't see how we can figure out what the eigenvalues would be anyway. I definitely don't see why the eigenvalues should be +/- 1. Any help would be great -- I just need help in seeing how this is ever the case. :grumpy: