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I'm reading a paper describing transformation of gradient of a vector $\mathbf u$ (velocity vector) when I came across the following: $\nabla \mathbf u = \mathbf q$ after transformation is, $$ \nabla_x \cdot (\mathbf u \otimes j \mathbf G^{-1}) = j\mathbf{q} $$ where, $u \in \mathbb R^3$, $\mathbf G$ is a $3\times 3$ matrix, j is a scalar (here, determinant of matrix $G$), $\mathbf q$ tensor or matrix (auxiliary variable role).

How do I go about expanding the tensor product?

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To simplify notation, we will let $\mathbf H$ stand for $j \mathbf G^{-1}$. Then the components of the tensor product $\mathbf u \otimes \mathbf H$ are

$\left(\mathbf u \otimes \mathbf H\right)^{ij}_k = \mathbf u^i \mathbf H^j_k$

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  • $\begingroup$ May I know in H^j_k what element does it stand for? H(j,k) or H(k,j) ? $\endgroup$
    – rationalrogu
    Mar 13, 2019 at 18:01
  • $\begingroup$ If you write vectors $\mathbf v^k$ as column vectors then in $\mathbf H^j_k$ we have $j$ as the row index and $k$ as the column index. Then with Einstein summation convention we have $(\mathbf Hv)^j=\mathbf H^j_k \mathbf v^k$. $\endgroup$
    – gandalf61
    Mar 14, 2019 at 9:11

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