Diagonalizing Hermitian matrices with adjoint representation

[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.

Thanks in advance

 

[fresh_42]

It is difficult to follow you. I think it will be clearer in a coordinate free wording. Complementations with zeros (or ones) are usually a hidden embedding. So what is here to be embedded where? And why are some $\lambda_a=0\,?$