I am reading an old paper [1] where they introduce a specific decomposition of a complex, invertible matrix. There is no proof, so I am trying to come up with one. The claim goes as follows:
Let $\mathbf{M} \in \mathrm{GL}_n(\mathbb{C})$ be an element of the general linear group (i.e. an $n \times n$, complex, invertible matrix). Then $\mathbf{M}$ can be written as \begin{equation} \tag{1} \mathbf{M} = \exp(\mathbf{m}) = \exp(\mathbf{m'}) \exp(\mathbf{m''}) \end{equation} where $\mathbf{m} \in \mathfrak{gl}_n(\mathbb{C})$ is an element of the general linear Lie algebra (i.e. an $n \times n$, complex matrix). The matrices $\mathbf{m}'$ and $\mathbf{m}''$ are $n \times n$, complex, block-matrices of the form \begin{align} \mathbf{m}' &= \begin{bmatrix} \mathbf{0}'_{00} & \mathbf{m}'_{01} & \mathbf{m}'_{02} \\ \mathbf{m}'_{10} & \mathbf{0}'_{11} & \mathbf{m}'_{12} \\ \mathbf{m}'_{20} & \mathbf{m}'_{21} & \mathbf{0}'_{22} \end{bmatrix}\tag{2} \\ \mathbf{m}'' &= \begin{bmatrix} \mathbf{m}''_{00} & \mathbf{0}''_{01} & \mathbf{0}''_{02} \\ \mathbf{0}''_{10} & \mathbf{m}''_{11} & \mathbf{0}''_{12} \\ \mathbf{0}''_{20} & \mathbf{0}''_{21} & \mathbf{m}''_{22} \end{bmatrix} \tag{3}. \end{align} The diagonal blocks are square and of matching dimensions ($\mathbf{0}'_{00}$ has the same dimensions as $\mathbf{m}''_{00}$, say $n_0 \times n_0$, and so on). In essence, $\mathbf{m}'$ has zero blocks on the diagonal while $\mathbf{m}''$ is block-diagonal. I realize that the number of blocks is not essential for the problem; I'm using three by three blocks for illustration.
I know that the exponential map of the general linear group is surjective, meaning that for every $\mathbf{M} \in \mathrm{GL}_n(\mathbb{C})$ there exists some $\mathbf{m} \in \mathfrak{gl}_n(\mathbb{C})$ such that $\mathbf{M} = \exp(\mathbf{m})$. So far so good. I can also see that block-diagonal matrices of the form (3) form a Lie algebra, say $\mathfrak{b}_n(\mathbb{C})$, which generates a Lie group of $n \times n$, complex, invertible, block-diagonal matrices, say $\mathrm{B}_n(\mathbb{C})$, which is a subgroup of $\mathrm{GL}_n(\mathbb{C})$. The paper suggests that the factorisation in (1) should be viewed as a coset decomposition. As far as I understand, the group $\mathrm{GL}_n(\mathbb{C})$ is the union of the left cosets \begin{equation} g \, \mathrm{B}_n(\mathbb{C}) = \{ g \, h \;|\; h \in \mathrm{B}_n(\mathbb{C}) \}, \quad g \in \mathrm{GL}_n(\mathbb{C}). \end{equation} I also know that some cosets may be identical so one doesn't need all cosets in order get the whole group (so to speak). What I don't understand is the specific form of the matrix $\mathbf{m}'$.
Any help would be much appreciated.
[1] J. Linderberg and Y. Öhrn, Int. J. Quantum Chem. 12(1), 161–191 (1977). State vectors and propagators in many-electron theory. A unified approach.