# Symmetry-breaking matrix/operator deformations: uniquely splitting eigenspaces into smaller ones?

Operators used in quantum mechanics, like Hamiltonian or angular momentum operator, usually have huge degeneracy of eigenspaces (symmetry inside them) - bringing a question of possibility to uniquely break this symmetry, e.g. can a deformation of matrix/operator split their eigenspaces? (determined by the matrix itself, not e.g. a random perturbation).

A natural intuition/answer is that it is impossible: standard matrix functions fulfill $$spec(f(A))=f(spec(A))$$ relation for eigenspectrum, can at most join eigenspaces if $$f$$ is not an injection.

However, recently working on graph isomorphism problem (stack), I got construction of kind of generalized matrix functions which, surprisingly, sometimes are able to reduce degeneracy, like splitting eigenspace into two smaller ones, for example:

$$t(A)_{ab}:=\sum_{ij} A_{ai} A_{aj} A_{ij} A_{ib} A_{jb}$$

It comes from "$$\triangleleft\hspace{-0.4mm}\triangleright$$"- shape graph with $$a, b$$ on left and right, $$i,j$$ on top and bottom. It is very strange matrix function, using kind of degree 3 matrix products: $$\sum_{a} M_{ab} M_{ac} M_{ad}$$. While standard matrix functions use standard degree 2 products, we can generalized them this way to defined by any graphs.

Examples of its reduction of eigenspectrum degeneracy come from adjacency matrices for strongly regular graphs (toughest cases for graph isomorphism problem, distinguished thanks to this function) - here are two 16x16 such $$\{0,1\}$$ coefficients symmetric matrices $$A$$ and $$B$$:

1111111000000000    1111111000000000
1111000111000000    1111000111000000
1111000000111000    1110100100110000
1111000000000111    1101010010001100
1000111100100100    1010101000101010
1000111010010010    1001011000010101
1000111001001001    1000111001000011
0100100111100100    0110000101010110
0100010111010010    0101000011101001
0100001111001001    0100001111000011
0010100100111100    0010100010111001
0010010010111010    0010010100110101
0010001001111001    0001100010101110
0001100100100111    0001010100011110
0001010010010111    0000101101001110
0001001001001111    0000011011110001


Eigenspectrum of both is the same: {7, 3, 3, 3, 3, 3, 3, -1, -1, -1, -1, -1, -1, -1, -1, -1}

Eigenspectrum of $$t(A)$$ is {145, 57, 57, 57, 57, 57, 57, 1, 1, 1, 1, 1, 1, 1, 1, 1}

Eigenspectrum of $$t(B)$$ is {145, 49, 49, 49, 49, 49, 49, 9, 9, 9, 9, 9, 9, 1, 1, 1}

Here is example of two 29x29 matrices having the same eigenspectrum with only 3 unique eigenvalues, such that $$t(A)$$ has still 3 unique eigenvalues, but $$t(B)$$ has all 29 unique eigenvalues - eigenspaces were completely split into 1D subspaces.

Replacing summation with integration in formula for $$t$$, we can get analogous deformation for operators - which could symmetry-break their eigenspaces.

Is there a literature for matrix/operator functions reducing eigenspectrum degeneracy?

## 1 Answer

can a deformation of matrix/operator split their eigenspaces?

Yes. Adding a random matrix $$X$$, $$A \mapsto A+\epsilon X$$, will lift any degeneracy.

Is there a literature for matrix/operator functions reducing eigenspectrum degeneracy?

Kato, Perturbation theory for linear operators.

• Ok, I should emphasize that it has to be unique - determined by the matrix/operator itself. Adding random matrix is not, above deformation 't' does it - sometimes can deform matrix having degenerated eigenspectrum, into matrix having nondegenerated eigenspectrum in an unique way - determined by the matrix/operator itself. – Jarek Duda Oct 30 '18 at 12:48
• I don't understand your comment – Norbert Schuch Oct 30 '18 at 17:47
• This symmetry breaking has to be determined by matrix/operator alone - you cannot add random matrix. A month ago I would say that it is impossible, but now I know examples - like written in the text above. – Jarek Duda Oct 30 '18 at 21:50