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)
 A: I suppose this matrix represents a Hamiltonian operator H. Transforming it into a block diagonal form means that we have exploited some symmetry respected by the system represented by this Hamiltonian to change the basis states used to represent H to a new set of states which obeys this symmetry, i.e., eigenstates of the symmetry operator. Each block then corresponds to a set of eigenstates of the symmetry operator having a certain eigenvalue.
A: First off, the unitary operator does not just involve mutliplying by phases, it also adds the rows and columns to each other. You really should think of the unitary operator as a rotation. It is just the generalization of a rotation for vectors or matrices with complex entires.
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.
