# Linear Algebra Tensor Proof

I want to ask for a general linear algebra formula that I found in the book Nonnegative Matrix and Tensor Factorizations in page 370, is that we can replace $\textbf{U}^{\otimes ^{-n}} \textbf{G}_{(n)}^T$ by $\overline{\textbf{X}}^T_{(n)} \textbf{U}^{(n)^{\dagger^T}}$

Specifying the rule with $N = 3$ and $n = 1$,

$(\textbf{U}^{(2)} \otimes \textbf{U}^{(3)}) \textbf{G}_{(1)}^T = \overline{\textbf{X}}^T_{(1)} \textbf{U}^{(1)^{\dagger^T}}$

I used this formula in an algorithm that wasn't converging except once every 10 times and then after doing this replacement it converged every single time.

Just for clarification this is for tucker decomposition,

$\cal{X} = \mathcal{G} \times_1 \textbf{U}^{(1)} \times_2 \textbf{U}^{(2)} \times_3 \textbf{U}^{(3)}$ for $N=3$ as $N$ is the order of the tensor $\mathcal{X}$, while $\mathcal{G}$ is the core tensor, $U^{(n)}$ are tensor factors.

$\textbf{U}^{(n)^T}$ is the transpose of nth factor matrix

$\times_n$ is the n-product

$\otimes$ is the kronecker product

$\dagger$ is the pseudo inverse

$\overline{\textbf{X}}_{(n)}$ over bar shows that this is the result of estimation of the unfolding n of the tensor $\mathcal{X}$

$\textbf{U}^{\otimes -n} = \textbf{U}^{(N)} \otimes ... \otimes \textbf{U}^{(n+1)} \otimes \textbf{U}^{(n-1)} \otimes ... \times \textbf{U}^{(1)}$

$\textbf{G}_{(n)}$ is the nth unfolding of the core tensor

Here is my question,

First I thought that both formulas would give the same result with difference in calculations cost. I need help how to confirm that both calculations are different or similar.

I replaced the numbers in the kronecker product $\textbf{U}^{(2)} \otimes \textbf{U}^{(3)} \neq \textbf{U}^{(3)} \otimes \textbf{U}^{(2)}$
The unfolding formula for Tucker $\textbf{X}_{(1)} = \textbf{U}^{(1)} \textbf{G}_{(1)} (\textbf{U}^{(3)} \otimes \textbf{U}^{(2)})^T$
then by using the formula in the book $\textbf{U}^{\otimes ^{-n}} \textbf{G}_{(n)}^T$ by $\overline{\textbf{X}}^T_{(n)} \textbf{U}^{(n)^{\dagger^T}}$
$\textbf{X}_{(1)} = \textbf{U}^{(1)} \textbf{U}^{(1)^{\dagger}} \overline{\textbf{X}}_{(1)}$ and because $\textbf{U}^{(n)}$ is orthogonal so it will come true.