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 lefthand side is greater than that of the righthand 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 $3\bigl(n^2{n+1\choose2}\bigr)+n=\frac12n(3n1)$. 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}?
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$\begingroup$ The lhs of (1) depends on $j$. Where is the dependence on $j$ on the rhs? $\endgroup$– Michael EngelhardtCommented Oct 6, 2022 at 5:47

$\begingroup$ @MichaelEngelhardt: Corrected. Thank you. $\endgroup$– HansCommented Oct 6, 2022 at 6:14

2$\begingroup$ Crossposted to math.SE: math.stackexchange.com/questions/4371111/… $\endgroup$– Qiaochu YuanCommented Oct 6, 2022 at 16:39
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1 Answer
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Asher Peres: Higher Order Schmidt Decomposition, Physics Letters, A 202, No. 1, 1617 (1995), MR1337627, Zbl 1020.81540, gives the necessary and sufficient condition.