# SVM without offset

I would like to know if the linear-SVM-without-offset solver: $$\min \frac{1}{2}\|w\|^2+C\sum_{i=1}^m \xi_i, \quad \mbox{s.t.}\quad y_iw^\top x_i \geq 1-\xi_i, \quad \xi_i\geq 0 \quad \forall i=1,\ldots,m.$$ can be applied to classify linearly separable data, where the hyperplane does not pass through the origin.

Maybe a change of coordinate system would help?

• Hi @Dikran. Thanks for the answer. Could you please be more specific about "adding an extra input feature..."? My thinking: Suppose that we have already an "SVMs-without-bias" solver $F(X,y,C)$ (that returns $w$ and $\xi$). Now for an instance $(X,y,C)$, where the data $(X,y)$ are linearly separable by a hyperplane that does not pass through the origin, if one applies a suitable rotation to the data to obtain $X',y'$ (i.e. we go to another space, the same idea as "feature space" in kernels), then the above solver can be used to classify the new (rotated) features: $F(X',y',C)$.
• If you just add an extra column of ones to $X$, then that will give an implicit bias parameter, even though you can still use the "SVMs-without-bias" solver. The extra element of $w$ will be the bias term. Apr 4, 2014 at 8:12