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I am new to SVM, but I would like to understand certain things.

Firstly, when dealing with multiclass classifications, I have a large number of support vectors as proven by R.

However, when I run svm.model in R

svm.model
We get the following:
Call:
svm(formula = churn ~ .,kernel = 'linear' , data = trainset, cost = 100, gamma = 1)
Parameters:
   SVM-Type:  C-classification
 SVM-Kernel:  linear
       cost:  100
gamma: 1
Number of Support Vectors: 598

However, a coworker who is extremely technical mentioned that there is always ONLY 1 support vector in a linear kernel. Am I missing something or these 598 support vectors can actually be combined into a single vector?

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I think you colleague meant that the result of linear SVM is one vector Q which defines if your point x belongs to class 1 or 0 (we are assuming two-class classification here) by looking if Q * x > 1 or not. Thus the decision boundary here is linear, hyperplane.

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  • $\begingroup$ Exactly. For linear SVM you will have more support vectors, just like kernel SVM, but they can be summarized as one vector in input space. Most packages dedicated to linear SVM do this (e.g. LIBLINEAR), but when you use a general SVM solver (e.g. LIBSVM) you typically end up with the support vectors and corresponding weights, even when using linear SVM. $\endgroup$ – Marc Claesen Jun 3 '15 at 13:34
  • $\begingroup$ How about class labels more than 2?? Sorry could you point to me any material that i could read more $\endgroup$ – aceminer Jun 3 '15 at 15:03
  • $\begingroup$ As I know multiclass classification is usually made by 1-vs-all method: for each class build own SVM to look for this particular class. If you want to know more about SVM there is good Coursera video course by Andrew Ng (class.coursera.org/ml-005/lecture/71 and 5 more videos). These videos are short and clear. $\endgroup$ – hvedrung Jun 3 '15 at 15:15

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