I am reading tensor decompositions and applications, by Tamara Kolda. There she mentions a property of

$\mathscr{Y} = \mathscr{X} \times_1 A^{(1)} \times_2 A^{(2)} \ldots \times_N A^{(N)} \iff$ $ Y_{(n)} = A^{(n)}X_{(n)}(A^{(N)}\otimes \ldots \otimes A^{(n+1)} \otimes A^{(n-1)} \otimes \ldots \otimes A^{(1)})^T$

Where $\mathscr{X},\mathscr{Y}$ are tensors, and $A^{(i)}$ is a matrix. $X_{(n)}$ is a mode $n$ matricized tensor $\mathscr{X}$.

Here $\otimes$ is a kronecker product and $\times_i$ represents $i$ mode product.

I am unable to find/do a proof of it. I tried it with examples, and obviously it worked. What is the proof of it, and what is the intuitive reasoning behind the statement?

For the definitions of these things, refer the hyperlink above.

This question has been asked before and is left unanswered.


I came up with this.

Let $X$ be $I_1\times\cdots \times I_N$ tensor and $A^{(n)}$ $J_n\times I_n$ matrices. By matricization definition, $y_{j_1\cdots j_N}$ element of $J_1\times\cdots\times J_N$ tensor $Y$ maps to $y_{j_nk}$ element of matrix $Y_{(n)}$, with $$ k=1+\sum_{\substack{l=1\\l\neq n}}^N (j_l-1)\prod_{\substack{m=1\\m\neq n}}^{l-1} J_m. $$ We will prove the statement by showing that every element $$ \left( X \times_1 A^{(1)} \times_2 A^{(2)}\times_3 \cdots \times_N A^{(N)}\right)_{j_1\cdots j_N} $$ is the same as element $$ \left[{A}^{(n)}{X}_{(n)}\left( {A}^{(N)} \otimes \cdots \otimes {A}^{(n+1)} \otimes {A}^{(n-1)} \otimes \cdots \otimes {A}^{(1)} \right)^T\right]_{j_nk}, $$ with $k$ as stated.

From $n$-mode product definition, we have $$ \left( {X} \times_1 {A}^{(1)} \times_2 {A}^{(2)}\times_3 \cdots \times_N {A}^{(N)}\right)_{j_1\cdots j_N}=\sum_{i_1=1}^{I_1}\cdots\sum_{i_N=1}^{I_N} x_{i_1\cdots i_N} a^{(1)}_{j_1i_1}\cdots a^{(N)}_{j_Ni_N}. $$ On the other hand, by denoting ${M}_n=\left( {A}^{(N)} \otimes \cdots \otimes {A}^{(n+1)} \otimes {A}^{(n-1)} \otimes \cdots \otimes {A}^{(1)} \right)^T$, we have \begin{align} \left({A}^{(n)}{X}_{(n)}{M}_n\right)_{j_nk}&={A}^{(n)}[\,j_n\,,\,:\,]\left({X}_{(n)}{M}_n\right)[\,:\,,\,k\,] \nonumber \\ &=\sum_{i_n=1}^{I_n} a^{(n)}_{j_ni_n}\left({X}_{(n)}{M}_n\right)[\,i_n\,,\,k\,] \nonumber \\ &=\sum_{i_n=1}^{I_n} a^{(n)}_{j_ni_n}\sum_{i=1}^{\hat{I}_n}{X}_{(n)}[\,i_n\,,\,i\,]{M}_n[\,i\,,\,k\,], \label{eq:proofjk} \end{align} with $\hat{I}_n=I_1\cdots I_{n-1}I_{n+1}\cdots I_N$. Now, ${X}_{(n)}[\,i_n\,,\,i\,]=x_{i_1\cdots i_N}$, with $$ i=1+\sum_{\substack{l=1\\l\neq n}}^N (i_l-1)\prod_{\substack{m=1\\m\neq n}}^{l-1} I_m. $$ The same $i$ stands in ${M}_n[\,i\,,\,k\,]$, which we easily get from the definition of Kronecker product $$ {M}_n[\,i\,,\,k\,]=\tilde{a}^{(N)}_{i_Nj_N}\cdots \tilde{a}^{(n+1)}_{i_{n+1}j_{n+1}}\tilde{a}^{(n+1)}_{i_{n-1}j_{n-1}}\cdots\tilde{a}^{(1)}_{i_1j_1}, $$ with $\tilde{a}^{(m)}_{i_mj_m}$ denoting element of ${{A}^{(m)}}^T$. Using these conclusions, we can rewrite the above as $$ \left({A}^{(n)}{X}_{(n)}{M}_n\right)_{j_nk}=\sum_{i_1=1}^{I_1}\cdots\sum_{i_N=1}^{I_N} x_{i_1\cdots i_N} a^{(1)}_{j_1i_1}\cdots a^{(N)}_{j_Ni_N}, $$ which completes the proof.

  • 1
    $\begingroup$ Great first answer! Well done! $\endgroup$ – user370967 Jun 6 '17 at 15:22
  • $\begingroup$ Thanks @Math_QED :) $\endgroup$ – Lana Periša Jun 7 '17 at 9:10

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