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.


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, $e^T\alpha = \langle e,\alpha \rangle = \sum_i \alpha_i$.

  • $\begingroup$ Thanks Yuval, just redone the maths and that interpretation works out. :) $\endgroup$ – user3450881 Apr 7 '14 at 16:47

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".

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.