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I'm studying multivariable calculus at the moment and have come across equations involving two bolded variables placed side by side, like so:

$ \nabla \mathbf{f}=\frac{\partial {{f}_{j}}}{\partial {{x}_{i}}}{{\mathbf{e}}_{i}}{{\mathbf{e}}_{j}} $

Is this meant to be a dot product? Or something else?

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    @Interested: $\mathbf{f}$ is a multivariable function $\mathbf{f}: \mathbb{R}^n \rightarrow \mathbb{R}^m$. The $j$ indexes the $j$-th component function of $\mathbf{f}$ or which we want to take the partial derivative wrt $x_i$.2011-05-21

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Since the result of $\nabla\mathbf{f}$ gives us a second-rank tensor (it can be represented as a two-by-two matrix with components $\frac{\partial f^j}{\partial x^i}$), the usual vector basis $\mathbf{e}_i=(0,\cdots,1,0,\cdots,0)$ (where the one is in the $i^{th}$ place) doesn't quite cut it. However, we can generalize them. If we define the matrices $\mathbf{e}_i\mathbf{e}_j$ to have all zeroes except for a one in the $(i,j)$ spot, then the sum $\sum_{i,j=1}^n \frac{\partial f^j}{\partial x^i} \mathbf{e}_i\mathbf{e}_j$ gives us the desired matrix, just as the sum $\sum_{i=1}^n \frac{\partial g}{\partial x^i} \mathbf{e}_i$ gives us $\nabla g$.

Properly, we should be writing $\mathbf{e}_i\bigotimes\mathbf{e}_j$ as we're actually dealing with the Kronecker product of $\mathbf{e}_i$ and $\mathbf{e}_j$, though in many contexts it seems people drop the direct sum symbol.

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    I am thinking of the canonical vectors as column vectors, so $\mathbf{e}_j\mathbf{e}_i^T$ is an $m \times n$ matrix. It is not a scalar. If you think of them as row vectors, then the product in question is $\mathbf{e}_j^T\mathbf{e}_i$, but that would be a very non-standard convention.2011-05-21