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I have a ($1 \times n$) row vector $\boldsymbol{x}$, an ($n \times n$) matrix $\mathbf{F}$, and an ($n \times n \times n$) tensor $\mathbf{Q}$. I also have a system of equations that reads

$\dot{x}_{i} = \sum\limits_{j=1}^{n}\sum\limits_{k=1}^{n} x_{j} x_{k} f_{jk} q_{jki}$,

where the $x$, $f$, and $q$'s correspond to elements of $\boldsymbol{x}$, $\mathbf{F}$, and $\mathbf{Q}$, respectively.

I would like to write the system of equations as a ($1 \times n$) row vector $\boldsymbol{\dot{x}}$ in terms of $\boldsymbol{x}$, $\mathbf{F}$, and $\mathbf{Q}$ given above, but I can't for the life of me figure out how.

Edit 1:

Ok, if I take $\mathbf{X} = \mathrm{diag}(\boldsymbol{x})$, then the sum over the $x_{j} x_{k}$ part can be described by $\mathbf{X}(\mathbf{X} + \mathbf{J}\mathbf{X}\mathbf{J})$, where $\mathbf{J}$ is the counter-identity matrix. Not sure if that is progress or not?

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2 Answers 2

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Typically, the Hadamard product is used where the objects on each side of the operand are the same size, e.g. $\,\,(X\circ Y)$ where $X,Y \in {\mathbb R}^{m\times n}$

So I would write this using the matrix product, the Hadamard product, and the double-dot product as $$\dot{x} = \big((xx^T)\circ F\big):Q$$ And since it is understood that the products can only be evaluated in one particular order, you can omit all of the parentheses.

NB: I know that you defined $x$ as a row vector, but $x^Tx$ looks like a scalar product to most readers, so I re-defined it as a column vector.

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  • $\begingroup$ Cool! I didn't actually know about the double-dot product. $\endgroup$ Commented Apr 29, 2015 at 9:02
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Using the Hadamard product I can write this as $\boldsymbol{x}(\mathbf{F} \circ \mathbf{Q}_{i})\boldsymbol{x}^{T}$.

I don't really like using the Hadamard product, so if anyone has a better idea, I would love to hear about it.

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