# Diagonalizing Hermitian matrices with adjoint representation

• A
Luck0
Suppose I have a hermitian ##N \times N## matrix ##M##. Let ##U \in SU(N)## be the matrix that diagonalizes ##M##: ##M = U\Lambda U^\dagger##, where ##\Lambda## is the matrix of eigenvalues of ##M##. This transformation can be considered as the adjoint action ##Ad## of ##SU(N)## over its algebra ##\mathfrak{su}(N)##, so if I consider the generators ##\{t_a\}## of ##\mathfrak{su}(N)##, I can write expansions: ##Ad(U)t_a = Ut_aU^\dagger = \sum_b\Omega_{ab}(U)t_b##, ##M = \sum_a m_at_a##, ##\Lambda = \sum_a \lambda_a t_a##, where ##\Omega_{ab} \in O(N^2-1)##, the orthogonal group, and the diagonalization can be written as ##M = \sum_{a,b}\lambda_b\Omega_{ab}t_a##.

Note that since ##\Lambda## is diagonal, the generators in the expansion ##\Lambda = \lambda_a t_a## must also be diagonal, i.e., they must span the Cartan subalgebra of ##\mathfrak{su}(N)##, meaning that some of the coefficients ##\lambda_a## must be zero. This means that some of the columns of each matrix ##\Omega## will not enter in the sum ##\sum_{a,b}\lambda_b\Omega_{ab}t_a##. My question is, if I look at ##M = \sum_{a,b}\lambda_b\Omega_{ab}t_a## as a change of coordinates, can I still consider ##\Omega## as an element of ##O(N^2-1)##? Clearly, if I write the elements of ##\Omega## that appear in the sum in matrix form, it will be a rectangular matrix, but I don't know if I can complete it with zeros until I get a square matrix, and if I can, I can't see what happens to orthogonality.