TITLE: Geometric meaning of block-diagonalization of a matrix
QUESTION [6 upvotes]: 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)

REPLY [2 votes]: 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.