I am looking at (the limitation of) the extension of the singular value decomposition to tensors. I would like to show that there is a tensor $A\_{i,j,k}$ that **cannot** be decomposed in the following singular value decomposition fashion
$$A\_{i,j,k}=\sum\_n \lambda\_n u\_{i,n}v\_{j,n}w\_{k,n} \tag 1\label{1}$$
where $\sum\_iu\_{i,m}^\*u\_{i,n}=\sum\_iv\_{i,m}^\*v\_{i,n}=\sum\_iw\_{i,m}^\*w\_{i,n}=\delta\_{m,n}$, $x^\*$ denotes the complex conjugate of $x$ and $\lambda\_n\ge 0$. Consider Equation \eqref{1} and confine ourselves to the complex number field. We can see the degree of freedom on the left-hand side is greater than that of the right-hand side whether in the complex number field or the real number field. For example, for a tensor $A$ of order $3$ and dimension $n$ in each component, the number of dimensions in $\mathbf R$ on the left hand side is $n^3$, whereas the number of dimensions in $\mathbf R$ on the right hand side is $3\bigl(n^2-{n+1\choose2}\bigr)+n=\frac12n(3n-1)$. The polynomial in $n$ of the left hand side has one higher degree than the right hand side. The critical value is $n=1$. Obviously there are $A$'s that cannot be decomposed as the right hand side. Is there a quick, intuitive way of finding one or even a systematic way of finding all the $A$'s that violates Equation \eqref{1}?



Singular value decomposition for tensor
[Asher Peres: *Higher Order Schmidt Decomposition*](https://arxiv.org/abs/quant-ph/9504006),
Physics Letters, A 202, No. 1, 16-17 (1995), [MR1337627](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1337627), [Zbl 1020.81540](https://zbmath.org/?q=an%3A1020.81540), gives the necessary and sufficient condition.


