Counterexample to Schmidt Decomposition for 3 systems Problem 2.77 in Nielsen and Chuang's book asks for an example of a pure state $|\psi\rangle$ in a three part composite space $A\otimes B\otimes C$ such that $|\psi\rangle$ has no Schmidt decomposition. 
So suppose I start with $|\psi\rangle = \sum_{i,j,k}c_{ijk}|i_A\rangle|j_B\rangle|k_C\rangle$ for orthonormal bases $|i_A\rangle, |j_B\rangle, |k_C\rangle$. The proof for 2 spaces involves taking the singular value decomposition of the coefficient matrix. So in the 3 space case, I could take the SVD of each matrix $C_k$ (defined as $(C_k)_{ij}=c_{ijk}$), but they might not all have the same SVD, and I think that causes the problem.
Using that, my best guess for a counterexample would be a state like 
$$|\psi\rangle = (|00\rangle + |01\rangle +|10\rangle+|11\rangle)|0\rangle + (|00\rangle+|01\rangle - |10\rangle - |11\rangle)|1\rangle$$
(appropriately normalized), since the parts of the vector corresponding to the first two spaces can't be singular value decomposed with the same unitaries (I think). But I have no idea how to prove this: How do I know there isn't some crazy basis for the space $C$ that would still produce a Schmidt decomposition?
 A: I did a more thorough literature search and found this paper giving necessary and sufficient conditions for a triple-Schmidt decomposition. 
So if a triple decomposition exists for $|\psi\rangle\in A\otimes B\otimes C$, then if I regard $B\otimes C$ as it's own vector space (call it $D$), then I ought to do the Schmidt decomposition for $A\otimes D$. So suppose I start with 
$$|\psi\rangle=\left(\frac{1}{\sqrt{3}}\right)|0\rangle \frac{|00\rangle+|11\rangle}{\sqrt{2}}+\left(\frac{\sqrt{2}}{\sqrt{3}}\right)|1\rangle\frac{|01\rangle+|10\rangle}{\sqrt{2}},$$
which is actually already in the Schmidt-decomposed form for $A\otimes D$. Since the coefficients are distinct for each term, then I think this decomposition is unique (at least, that's what it says in the linked paper). So if a Schmidt decomposition were to exist for the tripartite system, then I would have to express $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ as a single vector $|\phi\rangle\otimes|\psi\rangle$. This vector is entangled so this can't be done.
It seems like the trick is that as long as each vector and coefficient is distinct, the decomposition is unique. Our previous attempts failed because the coefficients were all equal. 
Thus, constructing a vector with a Schmidt decomposition in $A\otimes D$ that has such a form means we can "recurse" and try to Schmidt-decompose the resulting $B\otimes C$ vectors. Except when we do this, we will need to express $|\psi'\rangle\in B\otimes C$ as a single 
vector written as a tensor product $|\psi'_B\rangle\otimes|\psi'_C\rangle$, and it's easy to construct vectors where this can't be done.
