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I have a set of matrices $A_i$ that represent the generators of a finite group within a certain basis, and $B_i$ represent the same operators in a different basis.

How can I find a unitary transformation $U$ that performs this change of basis, so that $U A_i U^{-1} = B_i$ for all $i$? All I have are the matrices $A_i$ and $B_i$ and I need $U$.

For a single pair $A_1$ and $B_1$ it would be easy (as discussed here):

Let $P$ and $S$ be unitary matrices that diagonalize $A_1$ and $B_1$. Then $U A_1 U^{-1} = B_1$ is solved for $U=SP^{-1}$.

But this does not guarantee that the same $U$ will do the trick for all $A_i$ and $B_i$.

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    $\begingroup$ Why should $U = S P^{-1}$ be unitary? $\endgroup$
    – LSpice
    Commented May 2, 2021 at 21:42
  • $\begingroup$ @LSpice I think he is assuming $A_i$ and $B_i$ to be hermitian $\endgroup$
    – thedude
    Commented May 2, 2021 at 22:26
  • $\begingroup$ Because $S$ and $P$ can always be made unitary (since they are transformations that diagonalize $A$ and $B$). $\endgroup$
    – Rorsa
    Commented May 2, 2021 at 22:38
  • $\begingroup$ Sure, it can be arranged that $U$ (and $S$ and $P$) are unitary, but it is not automatic. (But neither is there such a thing as "the matrices that diagonalize"—indeed, the failure for there to be 'the matrix' is the whole point—so maybe it should be read as "some unitary matrices that diagonalize".) $\endgroup$
    – LSpice
    Commented May 2, 2021 at 22:47
  • $\begingroup$ Yes, all comments above are correct. And I'm sorry that I wasn't precise enough in the definitions and that I took some contextualization for granted. But I guess the problem is well established now. $\endgroup$
    – Gerson J Ferreira
    Commented May 2, 2021 at 23:10

1 Answer 1

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I've found a practical solution. But I still wonder if there's a more elegant and algebraic solution for this problem. In any case, here's my approach:

Let's consider that each $A_i$, $B_i$ and $U$ are $N\times N$ matrices, and there are $M$ pairs ($A_i, B_i$), i = 1...M. The proof that exists a common $U$ matrix that satisfies $U A_i U^{-1} = B_i$ for all $i$ follows from (finite) group theory, since $A_i$ and $B_i$ are simply different matrix representations of the same symmetry operator.

Start by linearizing (flatten) the matrix $U$ into a vector $\vec{U}$ in "Z"-order (line-by-line). Namely,

$$\vec{U}^T = [U_{11}, U_{12}, U_{13}, \cdots, U_{2,1}, U_{2,2}, \cdots U_{N,N}].$$

Now rewrite the set of equations $U A_i U^{-1} = B_i$ as $U A_i - B_i U = 0$, which can be cast in the linearized form as

$$Q \cdot \vec{U} = 0.$$ Here $Q$ is an $(N^2M) \times N^2$ rectangular matrix that can be written in terms of $N^2 \times N^2$ blocks for each $i$ as:

$$Q^T = [Q_1, Q_2, \cdots, Q_M],$$ and each block read as

$$Q_i = I \otimes A_i^T - B_i \otimes I.$$

Now, to find $\vec{U}$ we simply need to look for $Q$'s nullspace, or find the right SVD eigenvector with a zero singular value. This vector $\vec{U}$ can then be reshaped into the unitary matrix $U$.

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  • $\begingroup$ I think this can be seen in the light of the orbit-stabilizer theorem, where U is the unknown whose right orbit (acted by A) equals the left orbit (acted by B). $\endgroup$
    – daydreamer
    Commented May 2, 2021 at 22:35
  • $\begingroup$ If you want to speak of the eigenvector $\vec U$ (really the eigenline), then you'll want to require that the representation is irreducible. Otherwise there will be more than a line's worth of intertwiners, and the unitarity condition will be an extra quadratic (actually sesquilinear, because of the conjugation) constraint. $\endgroup$
    – LSpice
    Commented May 3, 2021 at 17:00
  • $\begingroup$ Indeed the unitary condition comes after. But in practice it is sufficient to find an $U$ that can be made unitary with a simple normalization, since I'm interested in the transformation itself, the determinant will always cancel in practice. I've implemented the code as described above and it seems to work well for the cases I've tested. It gives a single zero singular value. $\endgroup$
    – Gerson J Ferreira
    Commented May 3, 2021 at 23:24
  • $\begingroup$ You mean that I need $A_i$ and $B_i$ to be irreps? If they are not irreps, then I guess that there will be multiple $U$s, but they will differ only by a phase between the subspaces defined by the irreps. In practice I plan to apply this to the case where $A_i$ and $B_i$ are direct sums of irreps. So it should be ok. $\endgroup$
    – Gerson J Ferreira
    Commented May 3, 2021 at 23:24

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