I've been strugling with the following for a while. Will appreciate any direction.
If U is a unitary matrix and (U+iI) is hermitian, prove that U=-iI. (U is obviously square of order n).
Thanks,
SK
Think in terms of eigenvalues and eigenvectors. A unitary matrix is diagonalisable, so if all its eigenvalues are then U must be equal to . Also, each eigenvalue has absolute value 1.
If is hermitian then . Now suppose that is an eigenvalue of , with eigenvector x. Then and . Now over to you: show from those facts that .
Alternatively to [b]Opalg's[b] solution you can notice three things
1) The eigenvalues of are for any polynomial, thus the eigenvalues of are where
2) Since is Hermitian it's eigenvalues must be real, thus for each . Thus, for some
3) But, since is unitary we have that for every that and thus
Thus, all the eigenvalues of are .
Thanks for the idea. I think I understand the logic here. One thing that I'm missing is the jump from saying that if all the eigenvalues of U are -i then it is equal to -iI.
The only reasoning I can find is since U is unitary there exists an orthonormal base in which every single vector is an eigenvector, now if x is avector it can be represented as a linear combination of those base vectors and therfore one can show that U is kI (where k is the only eigenvalue).
BUT, I'm not sure that this is legitimate.
Thanks,
SK
Yes, that is a legitimate argument. Another way to see this result is to say that since there exists an orthonormal basis consisting of eigenvectors, the matrix of U with respect to that basis will be a diagonal matrix in which the diagonal elements are the corresponding eigenvalues. But if each of those eigenvalues is –i, then the matrix is –iI (where I is the identity matrix) and hence U = –iI (where I is the identity operator).
The easy part of the theorem I'm referring to says that if is an eigenvalue of and is a polynomial then is an eigenvalue of . This is easily verified since if is the associated eigenvector then clearly and thus
And we used it by noticing that if then so being an eigenvalue of implies that is an eigvenvalue of
Depending upon your knowledge this follows immediately since every unitary matrix is normal and thus we may appeal to the spectral theorem for normal matrices to conclude that where is unitary. Thus, noticing that we have proven that we see that implies that