# How to find a permutation matrix to turn a general hermitian matrix into a block diagonal one?

12 visualizaciones (últimos 30 días)
Huiyuan ZHENG el 24 de Abr. de 2022
Comentada: Huiyuan ZHENG el 30 de Jun. de 2022
If I have a hermitian matrix which satisfies H' = H, say H
H = [1, 0, -1i;
0, 1, 0;
1i, 0, 1],
is there a function to find a permutation matrix P so that
P'HP = [1, 0, 0;
0, 1, -1i;
0, 1i, 1]
a block diagonal one.
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

Christine Tobler el 18 de Mayo de 2022
First, we should keep in mind that the task is really to find a representation of A with as small blocks on the diagonal as possible. After all, we can always treat a matrix as block diagonal with one big block that's just the whole matrix.
One way you can think about this is to treat the matrix as an undirected graph: if A(i, j) is non-zero, there is an edge connecting node i and node j. We can then get the connected components of this graph (two nodes are in the same component only if there is a path connecting them). The nodes in each connected components represent the rows / columns of one diagonal block:
H = [1, 0, -1i;
0, 1, 0;
1i, 0, 1];
g = graph(H ~= 0);
plot(g)
nodeToComponent = conncomp(g) % nodeToComponent(i) gives component number of node i
nodeToComponent = 1×3
1 2 1
Why is there this mapping? If we try to split up a connected component into smaller blocks, this isn't possible because there is always an edge connecting a node in proposed new component A with another node in proposed new component B. In terms of diagonal blocks, that would mean that H(i, j) ~=0 for i in proposed diagonal block A and j in proposed diagonal block B - so this couldn't be seen as a diagonal block.
You can get a permutation vector from
[nodeToComponentSorted, p] = sort(nodeToComponent)
nodeToComponentSorted = 1×3
1 1 2
p = 1×3
1 3 2
H(p, p)
ans =
1.0000 + 0.0000i 0.0000 - 1.0000i 0.0000 + 0.0000i 0.0000 + 1.0000i 1.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 + 0.0000i
##### 1 comentarioMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos
Huiyuan ZHENG el 30 de Jun. de 2022
Dear Christine, thank you for your excellent answer! Decomposing matrices with the graph theory is really a very intuitive method.

Iniciar sesión para comentar.

### Más respuestas (1)

Bruno Luong el 24 de Abr. de 2022
A = [1, 0, 1;
0, 1, 0;
1, 0, 1],
A = 3×3
1 0 1 0 1 0 1 0 1
p=symrcm(A)
p = 1×3
3 1 2
A(p,p)
ans = 3×3
1 1 0 1 1 0 0 0 1
##### 3 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Bruno Luong el 25 de Abr. de 2022
Editada: Bruno Luong el 25 de Abr. de 2022
Your question is still not clear.
H is already block diagonal: a single block.
What is you criteria ? When I apply symrcm on you new small example of H, to me it is still fine:
H = [1, 0, -1i;
0, 1, 0;
1i, 0, 1];
p=symrcm(H);
Hp=H(p,p)
Hp-Hp' % still Hermitian?
% The below result shows that Hp have 2 Blocks: (2x2) and (1x1)
real(Hp)
imag(Hp)
Huiyuan ZHENG el 18 de Mayo de 2022
Sorry for the late response. I used symrcm to try my example matrix and it did work. However, when I tried to use symrcm to decompose a 14-by-14 hermitian matrix in my project, it was still not a block-diagonal one.
The problem may come that the 14-by-14 hermitian matrix is indeed not block-diagonalable.
Is there a proof that symrcm can always block-diagonalize a hermian matrix if it is block-diagonalable?
Actually, I want a function to help me decompose a representation matrix so that it can find the different irreducible representation and sort them into different blocks. Thus, the representation matrix can be arranged into different subspaces and each one corresponds to an irreducible representation.

Iniciar sesión para comentar.

### Categorías

Más información sobre Sparse Matrices en Help Center y File Exchange.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by