DownloadsDownload直積mitama2つのベクトルa,b\boldsymbol{a}, \boldsymbol{b}a,bのテンソル積a∘b=a⊗b=abT=(a1a2⋮an)(b1b2⋯bn)=(a1b1a1b2⋯a1bna2b1a2b2⋯a2bn⋮⋮⋱⋮anb1anb2⋯anbn)\boldsymbol{a} \circ \boldsymbol{b} = \boldsymbol{a} \otimes \boldsymbol{b} = \boldsymbol{a} \boldsymbol{b}^T = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \begin{pmatrix} b_1 & b_2 & \cdots & b_n \end{pmatrix} = \begin{pmatrix} a_1 b_1 & a_1 b_2 & \cdots & a_1 b_n\\ a_2 b_1 & a_2 b_2 & \cdots & a_2 b_n\\ \vdots & \vdots & \ddots & \vdots \\ a_n b_1 & a_n b_2 & \cdots & a_n b_n \end{pmatrix}a∘b=a⊗b=abT=⎝⎛a1a2⋮an⎠⎞(b1b2⋯bn)=⎝⎛a1b1a2b1⋮anb1a1b2a2b2⋮anb2⋯⋯⋱⋯a1bna2bn⋮anbn⎠⎞(1)を直積(direct product)あるいは外積(outer product)という。(例)A,B∈R2×2A, B \in \mathbb{R}^{2\times 2}A,B∈R2×2のとき、ATB=(a11a21a12a22)(b11b12b21b22)=(a11b11+a21b21a11b12+a21b22a12b11+a22b21a12b12+a22b22)\begin{align} A^T B &= \begin{pmatrix} a_{11} & a_{21}\\ a_{12} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{pmatrix} \\ &= \begin{pmatrix} a_{11} b_{11} + a_{21} b_{21} & a_{11} b_{12} + a_{21} b_{22}\\ a_{12} b_{11} + a_{22} b_{21} & a_{12} b_{12} + a_{22} b_{22}\\ \end{pmatrix} \end{align}ATB=(a11a12a21a22)(b11b21b12b22)=(a11b11+a21b21a12b11+a22b21a11b12+a21b22a12b12+a22b22)(3)であり、a1=(a11a12),b1T=(b11b12)a_1 = \begin{pmatrix} a_{11} \\ a_{12} \end{pmatrix} , \hspace{1em} b_1^T = \begin{pmatrix} b_{11} & b_{12}\\ \end{pmatrix}a1=(a11a12),b1T=(b11b12)(4)から∑i=12aibiT=(a11a12)(b11b12)+(a21a22)(b21b22)=(a11b11a11b12a12b11a12b12)+(a21b21a21b22a22b21a22b22)\begin{align} \sum^2_{i=1} \boldsymbol{a}_i \boldsymbol{b}_i^T &= \begin{pmatrix} a_{11} \\ a_{12} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \end{pmatrix} + \begin{pmatrix} a_{21} \\ a_{22} \end{pmatrix} \begin{pmatrix} b_{21} & b_{22} \end{pmatrix} \\ &= \begin{pmatrix} a_{11} b_{11} & a_{11} b_{12}\\ a_{12} b_{11} & a_{12} b_{12} \end{pmatrix} + \begin{pmatrix} a_{21} b_{21} & a_{21} b_{22}\\ a_{22} b_{21} & a_{22} b_{22} \end{pmatrix} \end{align}i=1∑2aibiT=(a11a12)(b11b12)+(a21a22)(b21b22)=(a11b11a12b11a11b12a12b12)+(a21b21a22b21a21b22a22b22)(5)であるため。