Tensor Composition_什么是 k-mode tensor-matrix product.-优快云博客

本文链接：https://blog.youkuaiyun.com/x5675602/article/details/104013563

本文探讨了张量与矩阵的k模式乘法，详细介绍了运算规则及特性。此外，还深入解析了三阶张量的Tucker分解，包括核心张量的概念及其通过单位矩阵进行j模式乘法的过程。

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Tensor Operator

Tensor times matrix: the k-mode product
The $k$ -mode product of a tensor $\boldsymbol{X} \in \mathbb{R}^{\boldsymbol{I}_{1} \times \boldsymbol{I}_{2} \times \ldots \times \boldsymbol{I}_{N}}$ with a matrix $\in \mathbb{R}^{J \times I_{k}}$ is written as
$\times_{k} A$
The resulting tensor $Y$ is of size $I_{1} \times \ldots \times I_{k-1} \times J \times I_{k+1} \times \ldots \times I_{N}$ , and contains the elements
$y_{i_{1} \cdots i_{k-1} j i_{k+1} \cdots i_{N}}=\sum_{i_{k}=1}^{I_{k}} x_{i_{1} i_{2} \cdots i_{N}} a_{j i_{k}}$
A few important facts about the k-mode product.
– $\times_m A \times_n B = X \times_n B \times_m A$ if $\neq m$
– but $\times_n A \times_n B = X \times_n (BA)$ (in general $\neq X \times_n B \times_n A$ )

Tucker Composition

For a 3rd-order tensor $\in F^{d_{1} \times d_{2} \times d_{3}}$ , where $F$ is either $\mathbb{R}$ or $\mathbb{C}$ , ‘’‘Tucker Decomposition’’’ can be denoted as follows,
$\mathcal{T} \times_{1} U^{(1)} \times_{2} U^{(2)} \times_{3} U^{(3)}$

where $\mathcal{T} \in F^{d_{1} \times d_{2} \times d_{3}}$ is the ‘‘core tensor’’, a 3rd-order tensor that contains the 1-mode, 2-mode and 3-mode singular values of $T$ , which are defined as the ''Frobenius norm" of the 1-mode, 2-mode and 3-mode slices of tensor $\mathcal{T}$ respectively. $U^{(1)}, U^{(2)}, U^{(3)}$ are unitary matrices in $F^{d_{1} \times d_{1}}, F^{d_{2} \times d_{2}}, F^{d_{3} \times d_{3}}$ respectively. Note that $\mathcal{T}$ might be much smaller than the original tensor $T$ if we accept an approximation instead of an exact equality. The CP decomposition can be seen as a special case of the Tucker decomposition, where the core tensor $\mathcal{T}$ is constrained to be superdiagonal.

The ‘‘j’’-mode product (’‘j’’ = 1, 2, 3) of $\mathcal{T}$ by $U^{(j)}$ is denoted as $\mathcal{T} \times U^{(j)}$ with entries as
$(\mathcal{T} \times_{1} U^{(1)})(d_{1}, d_{2}, d_{3}) = \sum_{i_{1}=1}^{d_{1}} \mathcal{T}(i_{1}, d_{2}, d_{3})U^{(1)}(d_{1}, i_{1})\\ (\mathcal{T} \times_{2} U^{(2)})(d_{1}, d_{2}, d_{3}) = \sum_{i_{2}=1}^{d_{2}} \mathcal{T}(d_{1}, i_{2}, d_{3})U^{(2)}(d_{2}, i_{2}) \\ (\mathcal{T} \times_{3} U^{(3)})(d_{1}, d_{2}, d_{3}) = \sum_{i_{3}=1}^{d_{3}} \mathcal{T}(d_{1}, d_{2}, i_{3})U^{(3)}(d_{3}, i_{3})$