dot product between vector and matrix

In my book on fluid mechanics there is an expression $$\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij}$$ where $\boldsymbol{\tau}_{ij}$ is a rank-2 tensor (=matrix). Given that $\boldsymbol\nabla=(\partial_x, \partial_y, \partial_z)$, a vector, what do I get when I dot it with a matrix?

If I was to write $\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij}$ in Einstein notation, then how would it look?

The tensor $\boldsymbol{\tau}_{ij}$ is given by $$\begin{pmatrix} \tau_{xx} & \tau_{yx} & \tau_{zx} \\ \tau_{xy} & \tau_{yy} & \tau_{zy} \\ \tau_{xz} & \tau_{yz} & \tau_{zz} \end{pmatrix}$$ and the dot-product yields (by comparison with later expressions in the chapter)

$$\boldsymbol{\nabla}\cdot \boldsymbol{\tau}_{ij} = \mathbf{i}(\partial_x \tau_{xx} + \partial_y \tau_{yx} + \partial_z \tau_{zx})+\mathbf{j}(\partial_x \tau_{xy} + \partial_y \tau_{yy} + \partial_z \tau_{zy}) + \mathbf{k}(\partial_x \tau_{xz} + \partial_y \tau_{yz} + \partial_z \tau_{zz})$$

However, I don't see how this last expression comes about.

• The only reasonable motivation to use a dot here is to indicate $\partial_j \tau_{ij}$ (instead of $\nabla \tau_{ij} = \partial_i \tau_{ij}$), but I have not seen this notation before. – Phira Dec 29 '13 at 21:58
• I think Phira is correct. But the book really messed up with its notation. I should either write what Phira wrote, or write $\nabla \cdot \tau$. The result should be a vector. – Stephen Montgomery-Smith Dec 29 '13 at 22:07
• @Phira thanks for the comments, I have added more info to the question. maybe that helps? – BillyJean Dec 29 '13 at 22:14
• Your expression corresponds to what I wrote. It looks swapped because the indices in your matrix are swapped compared to the usual convention. You should imagine the $\nabla$ to be a row vector that is multiplied with the usual dot product with the first row of the matrix to give the first component of the resulting vector (Which is the coefficient of your $\bf i$). – Phira Dec 29 '13 at 22:31
• @Phira thanks -- don't you mean that the row from $\nabla$ should be multiplied by each column of $\tau$? – BillyJean Dec 29 '13 at 22:34