# There are $n!$ different unitary similarities that give $L_i \sim L_j$. Or are there more?

Given a matrix $L$ of dimension $n \times n$ which is lower triangular, and with distinct elements along the diagonal, there is a way to unitarily transform it into a different lower triangular matrix as follows.

With $L$'s lower triangular, $P$'s permutations, $D$'s diagonal, $Q$'s unitary, and $C$'s non-sungular: \begin{align} CL_0C^{-1}&=D_0 \\ PD_0P^*&=D_1 \\ QD_1Q^* &= L_1 \\ \end{align}

The claim is that the total change is a unitary similarity, or equivalently that $L_1 = UL_0 U^*$ for some $U$ unitary.

Begin Placeholder On the likely chance that clarifications of my claim are necessary, I will insert/edit here later. For now a (hopefully complete and clear) sketch.

The matrix is first diagonalized and re-ordered via permutation so that we have : $$D_1=PCL_0C^{-1}P^*$$

The original matrix $L_0$ is now represented in a similar diagonal form with respect to the re-ordered diagonal matrix $D_1$: $$L_0 = C^{-1}\underbrace{P^*D_1P}_{D_0}C$$

From this point, similarity operations may be applied to $D_1$ putting it in lower triangular form. This is done with the goal of the unitary $U$ in mind:

Factor $C^{-1}P^*=UR$ from the QR factorization of $C^{-1}P^*$ where $U$ is unitary. Then $$L_0 = \underbrace{UR}_{C^{-1}P^*}D_1R^{-1}U^*$$

Since $R$ is lower triangular, its inverse is also, and we have $L_1 = RD_1R^{-1}$ is lower triangular. Thus the unitary similarity is shown $$L_0 = UL_1 U^*$$

End Placeholder

This all amounts to re-ordering the diagonal elements of $L_i$ while unitarily retaining the lower triangular structure. Thus since there are $n!$ such possible orderings resulting in distinct $L_i$, there are $n!$ different unitary transformations. So for my question:

Does this enumerate (up to the number of distinct diagonal elements) all such unitary similarities between lower triangular forms?

I don't know the answer to this question. I am very interested in any discussion since differing viewpoints will likely be enlightening.

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$Q$ and $P$ are unitary. Since the unitary matrices form a group, $QPC$ is unitary if and only if $C$ is unitary.