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I'm trying to understand the material in "A Dual Coordinate Descent Method for Large-scale Linear SVM" by Hsieh et. al. (link to paper) There is an equation for the Dual form of an unconstrained optimisation problem,

$$ f(\mathbf{\alpha})=\dfrac{1}{2}\mathbf{\alpha}^T\bar{Q}\alpha-e^T\alpha $$

I don't understand what the $\mathbf{e}^T$ means, it's not explained in the surronding text, so I assume it's just some common notation. Later in the paper $\mathbf{e}_i$ is defined as $\mathbf{e}_i=[0,\ldots,1,0,\ldots,0]^T$, so maybe it's some sort of selector term? Not sure if this second mention is even related.

Please may someone explain to be what the $\mathbf{e}^T$ bit is doing in this formula? Thank you for your time.

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The superscript $T$ is the transpose operation. The vector $e$ is probably the constant one vector, though this is not standard notation and so it would be mentioned in the definitions part of the paper. Overall, $e^T\alpha = \langle e,\alpha \rangle = \sum_i \alpha_i$.

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  • $\begingroup$ Thanks Yuval, just redone the maths and that interpretation works out. :) $\endgroup$ – user3450881 Apr 7 '14 at 16:47
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Although it has already been answered, I would just like to point out that the answer is definitely correct. On the bottom left column of page 12 the authors state "$e$ is the vector of ones".

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  • $\begingroup$ Ah, thanks for that. That's what I get for skimming proofs! $\endgroup$ – user3450881 Apr 8 '14 at 13:38

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