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

조회 수: 9(최근 30일)
Huiyuan ZHENG
Huiyuan ZHENG 2022년 4월 24일
답변: Christine Tobler 2022년 5월 18일
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.

답변(2개)

Bruno Luong
Bruno Luong 2022년 4월 24일
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
  댓글 수: 3
Huiyuan ZHENG
Huiyuan ZHENG 2022년 5월 18일
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.

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Christine Tobler
Christine Tobler 2022년 5월 18일
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

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