Two TT truncations realize the root-sum-square bound

With columns ordered by $(i_2,i_3)=(1,1),(1,2),(2,1),(2,2)$, the first unfolding is $\left(\begin{array}{cccc}3& 0& 0& 1\\ 0& 2& 0& 0\end{array}\right)$. Its rows are orthogonal with norms $\sqrt{10}$ and $2$, so the leading singular value is $\sqrt{10}$ and the first discarded tail is ${\delta }_1=2$. The retained left vector is $(1,0{)}^{\mathsf{T}}$, and the remainder ${\mathrm{\Sigma }}_1{V}_1^{\ast }$ reshaped as a $2\times 2$ matrix is $\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right)$.

The second remainder has singular values $3$ and $1$, hence its rank-one truncation discards ${\delta }_2=1$ and returns the tensor $\widehat{A}=3e_1\otimes e_1\otimes e_1$. The omitted entries of $A-\widehat{A}$ are $e_1\otimes e_2\otimes e_2$ and $2e_2\otimes e_1\otimes e_2$, which are orthogonal and have squared norms $1$ and $4$. Therefore $\mathrm{\parallel }A-\widehat{A}{\mathrm{\parallel }}_F=\sqrt{5}=\sqrt{{\delta }_1^2+{\delta }_2^2}$, realizing equality in [F3].