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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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Or the multiplication of a row vector with a column vector? But that doesn't work.. –  Josh Chen May 20 '11 at 23:45
    
I suppose it would make sense if $\mathbf{f}$ was defined on one of these or one of these. –  Ricky Demer May 20 '11 at 23:57
    
@Josh: Could you provide the context for $\mathbf{f}$? The 'j' subscript in particular. –  InterestedGuest May 20 '11 at 23:58
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It looks like (since the f on the left it bold) that they're taking the gradient of a vector (which is properly called the covariant derivative). If that's the case, then $\mathbf{e}_i\mathbf{e}_j$ is a matrix with all zeroes except for a 1 in the $(i,j)$th slot (a generalization of the unit vectors). The $f_j$ are the component functions of $\mathbf{f}$, and it looks like there's an implicit summation on both $i$ and $j$. Essentially, you're taking the gradient of each of the component functions and putting those as the entries of a vector (it's not quite that simple, but that's the gist). –  Riley E May 21 '11 at 0:55
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@Riley: $\nabla\mathbf{f}$ is indeed the gradient of the vector function $\mathbf{f}$, so you've got the context right. My apologies for not making it clear. It was the $\mathbf{e}_i\mathbf{e}_j$ I wasn't sure about. You say it's a single notation for a matrix? Could you put that into an answer so I can accept it if you don't mind? –  Josh Chen May 21 '11 at 3:08
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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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Something like $\mathbf{e}_i\mathbf{e}_j^T$ would be more explicit, and hardly more cumbersome. –  yasmar May 21 '11 at 14:11
    
Except I have it backwards. Apparently in this context, $\mathbf{e}_i\mathbf{e}_j$ means $\mathbf{e}_j\mathbf{e}_i^T$, where $e_i$ is a canonical basis vector for $\mathbb{R}^n$, $e_j$ for $\mathbb{R}^m$, and $f:\mathbb{R}^n \to \mathbb{R}^m$. –  yasmar May 21 '11 at 14:24
    
Actually, $\mathbf{e}_i\mathbf{e}_j^T$ would give a scalar which we don't want. If $A$ and $B$ are two vectors, then $(A\bigotimes B)_{ij} = A_i B_j$. It's literally the product of every component of $A$ with every component of $B$. –  Riley E May 21 '11 at 15:11
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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. –  yasmar May 21 '11 at 17:34
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