# Is an SVM the same as a neural network without a hidden layer?

A neural network without a hidden layer is the same as just linear regression.

If I then use squared hinge loss and encoporate the l2 regularisation term, is it fair to then call this network the same as a linear SVM?

Going by this assumption, then if I need to implement a multiclass SVM, i can just have n output nodes (where n is the number of classes). Would this then be equivalent to having n number of SVMs, similar to a one-vs-rest method?

If I then wanted to encoporate a kernel into my SVM, could I then use an activation function or layer prior to the final output nodes (where I compute loss and add regularisation) which would then transfer this data into another feature plane the same as that of an SVM kernel?

This is my current hunch, but would like some confirmation or correction where my understanding is incorrect.

• "A neural network without a hidden layer is the same as just linear regression.", this is not true. This is only true if the activation functions are identities. So you'd better review your post and question. – nbro Apr 27 at 20:26

First, what makes the neural network different than linear regression is the non-linearity (acivation function), not the number of layers. So, a neural network with $$n$$ layers with no non-linearities is still the same as linear regression. Second, SVM finds the hyperplane of maximum margin. You are not guaranteed to find that hyperplane that has maximum margin with a neural network (using non-linearities).