Singular value decomposition for tensor

Question

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 $$A_{i,j,k}=\sum_n \lambda_n u_{i,n}v_{j,n}w_{k,n} \tag 1$$ 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 $(1)$ 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, on the left-hand side, for a tensor $A$ of order $3$ and dimension $n$ in each component, the degree of freedom in the real number is $n^3$. However, on the right hand side, the degree of freedom is $3\big(n^2-{n+1\choose2}\big)+n=\frac12n(3n-1)$. 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 $(1)$?

Answer 1

Asher Peres, Higher Order Schmidt Decomposition provides the necessary and sufficient condition.