Before I know how to do tucker decomposition, I mistakenly thought the core tensor is only from combining the singular value matrices of the flattening matrices. Yes I know it is not now. For the tucker decomposition method for a 3 way data's tensor decomposition, a core tensor is calculated using the original tensor(e.g. 3D tensor) and its 3 flattening matrices' left singular vectors. I am wondering if there is any relation existing between the singular values of the flattening matrices and the core tensor somehow? Just out of intuition.

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Here are the materials about Tucker decomposition for tensors:

  1. http://en.wikipedia.org/wiki/Tucker_decomposition
  2. Tensor Decompositions and Applications
  • $\begingroup$ It would help non-specialist readers of your question if you defined or linked to the Tucker decomposition and flattening matrices. $\endgroup$
    – Bill Barth
    Commented Apr 22, 2014 at 11:38

1 Answer 1


First, an important distinction must be done. A Tucker decomposition of a third-order tensor $\mathcal{T}$ is any decomposition of the form $$ \mathcal{T} = \mathcal{G} \times_1 \mathbf{A} \times_2 \mathbf{B} \times_3 \mathbf{C}, $$ where $\mathcal{G}$ is also a third-order tensor called core and $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are the matrix factors of the decomposition. In scalar form, $$ t_{i_1,i_2,i_3} = \sum_{r_1=1}^{I_1} \sum_{r_2=1}^{I_2} \sum_{r_3=1}^{I_3} g_{r_1,r_2,r_3} a_{i_1,r_1} b_{i_2,r_2} c_{i_3,r_3}. $$ where $I_1, I_2, I_3$ are the dimensions of $\mathcal{T}$. The flat (horizontal) matrix unfoldings of $\mathcal{T}$ are then written as $$ \mathbf{T}_1 = \mathbf{A} \mathbf{G}_1 (\mathbf{C} \otimes \mathbf{B})^T \\ \mathbf{T}_2 = \mathbf{B} \mathbf{G}_2 (\mathbf{A} \otimes \mathbf{C})^T \\ \mathbf{T}_3 = \mathbf{C} \mathbf{G}_3 (\mathbf{B} \otimes \mathbf{A})^T, $$ where $\otimes$ denotes the Kronecker product.

When the factor matrices are the matrices of left singular vectors of the singular value decompositions of each matrix unfolding, then we have a particular Tucker decomposition which is usually called the higher-order singular value decomposition (HOSVD) (see the seminal paper by De Lathauwer et al., ftp://ftp.esat.kuleuven.be/pub/SISTA/delathauwer/reports/ldl-94-31.pdf). Mathematically, expressing the SVDs of $\mathbf{T}_1$, $\mathbf{T}_2$ and $\mathbf{T}_3$ as $$ \mathbf{T}_n = \mathbf{U}^{(n)} \mathbf{\Sigma}^{(n)} {\mathbf{V}^{(n)}}^H, \quad n = 1,2,3, $$ the HOSVD of $\mathcal{T}$ is given by $$ \mathcal{T} = \mathcal{S} \times_1 \mathbf{U}^{(1)} \times_2 \mathbf{U}^{(2)} \times_3 \mathbf{U}^{(3)}, $$ where the core satisfies $$ \mathcal{S} = \mathcal{T} \times_1 {\mathbf{U}^{(1)}}^H \times_2 {\mathbf{U}^{(2)} }^H \times_3 {\mathbf{U}^{(3)}}^H. $$ It is easy to verify the above formula for the core, as $$ \mathbf{S}_1 = {\mathbf{U}^{(1)}}^H \mathbf{T}_1 ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T \\ = {\mathbf{U}^{(1)}}^H \mathbf{U}^{(1)} \mathbf{S}_1 ({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}})^T ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T = \mathbf{S}_1 $$ because each $\mathbf{U}^{(n)}$ is unitary by definition.

Now, the answer to your question can be immediately taken from the above formulas, since substituting the SVD of $\mathbf{T}_1$ in the expression for $\mathbf{S}_1$ we have $$ \mathbf{S}_1 = {\mathbf{U}^{(1)}}^H \mathbf{T}_1 ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T = {\mathbf{U}^{(1)}}^H \mathbf{U}^{(1)} \mathbf{\Sigma}^{(1)} {\mathbf{V}^{(1)}}^H ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T = \mathbf{\Sigma}^{(1)} {\mathbf{W}^{(1)}}^H $$ where we have defined $$ {\mathbf{W}^{(1)}} = ({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}})^T {\mathbf{V}^{(1)}}. $$ Note that $$ {\mathbf{W}^{(1)}}^H {\mathbf{W}^{(1)}} = {\mathbf{V}^{(1)}}^H ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)^T ({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}})^T {\mathbf{V}^{(1)}} \\ = {\mathbf{V}^{(1)}}^H [({\mathbf{U}^{(3)}} \otimes {\mathbf{U}^{(2)}}) ({\mathbf{U}^{(3)}}^H \otimes {\mathbf{U}^{(2)}}^H)]^T {\mathbf{V}^{(1)}} \\ = {\mathbf{V}^{(1)}}^H {\mathbf{V}^{(1)}} = \mathbf{I}, $$ and hence $\mathbf{W}^{(1)}$ is unitary. This tells us that the mode-1 unfolding of the core $\mathcal{S}$ is a product of the diagonal matrix of singular values of $\mathbf{T}_1$ by a unitary matrix. Therefore, the norm of the $i$th row of $\mathbf{S}_1$ is precisely the $i$th singular value $\sigma_i^{(1)}$ of $\mathbf{T}_1$. An analogous reasoning applies of course to the other modes.

Finally, observe also that $$ \|\mathcal{S}\|_F^2 = \|\mathbf{S}_n\|_F^2 = \text{Trace}(\mathbf{S}_n\mathbf{S}_n^H) = \sum_i (\sigma_i^{(n)})^2 = \|\mathbf{T}_n\|_F^2 = \|\mathcal{T}\|_F^2,$$ which comes as no surprise, since $\mathcal{S}$ is given by a unitary multilinear transformation of $\mathcal{T}$.


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