# Geometric meaning of block-diagonalization of a matrix

some times we need to do block-diagonalization in favor of easy computation. For instance, for a matrix like this

$$\begin{bmatrix} A_{11} & A_{12} & A_{13} & 0 & 0 & 0\\ A_{21} & A_{22} & A_{23} & 0 & 0 & c\\ A_{31} & A_{32} & A_{33} & 0 & c & 0\\ 0 & 0 & 0 & B_{11} & B_{12} & B_{13}\\ 0 & 0 & c & B_{21} & B_{22} & B_{23}\\ 0 & c & 0 & B_{31} & B_{32} & B_{33} \end{bmatrix}$$

we want to block-diagonalize it, i.e. eliminate 'c' by merging it into the block-diagonal part, we can do this by using some transformation matrix(unitary operator here) which only involves some phase terms $me^{i\theta}$, I would like to know what is the geometric meaning of this operation?(seems not like a usual rotation)

Assuming this operator is your hamiltonian $H$, making this transformation changes the problem so you can now think of it as two independent $3$ dimensional systems instead of one $6$ dimensional system. This is because the propagator of the the system, $U(t) = e^{iHt}$ will also be block diagonal, so if a state starts out with non-zero components in only the top three components, it will remain that way. Same for the bottom three components.