# norm of SVM's weights vector

From solving a hard-margin SVM primal problem we get: $$w = \sum{\alpha_i y_i x_i} \\ \sum{a_i y_i} = 0$$

Where $$\alpha$$ is the lagrangian multiplier vector. After solving for $$w$$ (using the dual problem) we can also solve for $$b$$ (the bias) with one of the support vectors by $$y_i(wx_i+b)=1$$

Given that, I'm looking for a way to prove that $$\lVert w \rVert ^2 = \sum{\alpha_i}$$

Thanks.

Multiply both sides of the third expression with $$\alpha_i$$ and sum $$\forall i$$: $$A=\sum_i \alpha_iy_i(w^Tx_i+b)=\sum_i\alpha_i$$ RHS is what we want, the LHS can be evaluated as: $$A=w^T\left(\sum_i{\alpha_i y_i x_i}\right)+b\sum_i{\alpha_i y_i}=w^Tw+0=||w||^2$$