Singular value decomposition for tensor

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}?

• The lhs of (1) depends on $j$. Where is the dependence on $j$ on the rhs? Commented Oct 6, 2022 at 5:47
• @MichaelEngelhardt: Corrected. Thank you.
– Hans
Commented Oct 6, 2022 at 6:14
• Crossposted to math.SE: math.stackexchange.com/questions/4371111/… Commented Oct 6, 2022 at 16:39