By assumptions a and c and because SS = I , The final line is an eigendecomposition of G = S ( J ab ) − $^{1}$SJ ab whose left-eigenvectors are the rows of P $^{ma}$. The eigenvectors of G are unique by assumption b, so the eigendecomposition is unique up to scaling of the eigenvectors. As the rows of P ma must sum to one, these scaling factors are uniquely determined. We have shown that the rows of P ma can be uniquely recovered from J $^{ab}$. The order of these rows is not identifiable without further assumptions, so for the moment we will say that P ma can be identified up to a set of one of eight permutations of the rows such that the resulting P ma is strand-symmetric in form. Then, by assumption c, π m = 1 J $^{ab}$( P $^{ma}$) − 1 and ( P mb ) ⊺ = J $^{ab}$(Π$^{m}$P $^{ma}$) − $^{1}$, so π m and P mb are also identifiable up to one of eight permutations of their elements and rows that correspond to the permutation chosen for P $^{ma}$. In this section we build on a result from Chang (1996) to demonstrate that, under mild conditions, the rooted tree topology is identifiable under a non-stationary, strand-symmetric Markov model from the pairwise distributions of states at terminal nodes. We first need to also import an equivalence relation between tree topologies. Let T$_{1}$ = ( S$_{1}$,E$_{1}$ ) and T$_{2}$ = ( S$_{2}$,E$_{2}$ ) be trees with the same set of terminal nodes T . We say that T$_{1}$ and T$_{2}$ are equivalent if there is a bijective “relabelling” function from S$_{1}$ to S$_{2}$ that is the identity function for the terminal nodes and such that the edges 3.2 Identifiability of the Topology 11 S ( J $^{ab}$) − $^{1}$SJ ab = S ( P $^{ma}$) − $^{1}$SS (Π $^{m}$) − $^{1}$(( P $^{mb}$) $^{⊺}$) − $^{1}$S ( P $^{mb}$) $^{⊺}$SS Π $^{m}$P ma = ( P $^{ma}$) − $^{1}$S (Π $^{m}$) − $^{1}$S Π $^{m}$P ma = ( P $^{ma}$) − $^{1}$diag ( π m (1) π m (4) π m (2) π m (3) π m (3) π m (2) π m (4) π m (1) ) P $^{ma}$.