I am reading this article "Tensor Decompositions and Applications" by Kolda and Bader. On page 21, it says:
...CP [decomposition] can be viewed as a special case of Tucker [decomposition] where the core tensor is superdiagonal and $P = Q = R$...
But there's no proof to back the statement. Am I missing something here?
The proof in (Kolda and Bader, 2009) may be left out since the authors consider it trivial.
(CP decomposition) Let $\textbf{X} \in \mathbb{R}^{I \times J \times K}$ be a 3-way tensor. CP decomposition of $\textbf{X}$ can be written elementwise as $$ \textbf{X} \approx \sum_{r = 1}^{R} a_{ir} b_{jr} c_{kr}, ~~~ \text{for} ~~~ i = 1,\ldots,I, ~~ j = 1,\ldots,J, ~~ k = 1,\ldots,K, $$ where $R$ is a positive integer, $\textbf{a}_r \in \mathbb{R}^{I}$, $\textbf{b}_r \in \mathbb{R}^{J}$, and $\textbf{c}_r \in \mathbb{R}^{K}$ for $r = 1,\ldots,R$.
(Tucker decomposition) Let $\textbf{X} \in \mathbb{R}^{I \times J \times K}$ be a 3-way tensor. Tucker decomposition of $\textbf{X}$ can be written elementwise as $$ \textbf{X} \approx \sum_{p = 1}^{P} \sum_{q = 1}^{Q} \sum_{r = 1}^{R} g_{pqr} a_{ip} b_{jq} c_{kr}, ~~~ \text{for} ~~~ i = 1,\ldots,I, ~~ j = 1,\ldots,J, ~~ k = 1,\ldots,K, $$ where $P,Q,R$ is a positive integer, $\textbf{A} \in \mathbb{R}^{I \times P}$, $\textbf{B} \in \mathbb{R}^{J \times Q}$, and $\textbf{C} \in \mathbb{R}^{K \times R}$, and $\textbf{G} \in \mathbb{R}^{P \times Q \times R}$ is the core tensor.
(Superdiagonal tensor) A tensor $\textbf{X} \in \mathbb{R}^{I_1 \times I_2 \times \ldots \times I_N}$ is superdiagonal if $$ x_{i_1 i_2 \ldots i_N} = \begin{cases} 1, & ~~\text{if} ~~ i_1 = i_2 = \ldots = i_N \\ 0, & ~~\text{otherwise} \end{cases} $$
Statement. CP decomposition can be viewed as a special case of Tucker where the core tensor is superdiagonal and $P = Q = R$.
Proof. The proof uses simple algebra to show that an arbitrary Tucker decomposition with superdiagonal core tensor can be reduced to a CP decomposition. Let $[\![\textbf{G}; \textbf{A}, \textbf{B}, \textbf{C}]\!]$ be a Tucker decomposition that satisfies $$ \tilde{\textbf{X}} = \sum_{p = 1}^{P} \sum_{q = 1}^{Q} \sum_{r = 1}^{R} g_{pqr} a_{ip} b_{jq} c_{kr}, ~~~ \text{for} ~~~ i = 1,\ldots,I, ~~ j = 1,\ldots,J, ~~ k = 1,\ldots,K, $$ where $\textbf{G}$ is superdiagonal and $\tilde{\textbf{X}} \in \mathbb{R}^{I \times J \times K}$ be a 3-way tensor. Then this equation can be rewritten as \begin{align} \tilde{\textbf{X}} & = \sum_{p = 1}^{P} \sum_{q = 1}^{Q} \sum_{r = 1}^{R} g_{pqr} a_{ip} b_{jq} c_{kr} \\ & = \sum_{p = 1}^{P} \left( \sum_{q = 1, q \neq p}^{Q} \sum_{r = 1}^{R} g_{pqr} a_{ip} b_{jq} c_{kr} + \sum_{r = 1}^{R} g_{ppr} a_{ip} b_{jp} c_{kr} \right) \\ & = \sum_{p = 1}^{P} \sum_{r = 1}^{R} g_{ppr} a_{ip} b_{jp} c_{kr} \\ & = \sum_{p = 1}^{P} \left( \sum_{r = 1, r \neq p}^{R} g_{ppr} a_{ip} b_{jp} c_{kr} + g_{ppp} a_{ip} b_{jp} c_{kp} \right) \\ & = \sum_{p = 1}^{P} a_{ip} b_{jp} c_{kp}. \end{align} Then, $[\![\textbf{A}, \textbf{B}, \textbf{C}]\!]$ is a CP decomposition of $\tilde{\textbf{X}}$.
Kolda, T. G., & Bader, B. W. (2009). Tensor decompositions and applications. SIAM review, 51(3), 455-500. doi: 10.1137/07070111X
