norm of SVM's weights vector

Question

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.

Answer 1

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: $$A=w^T\left(\sum_i{\alpha_i y_i x_i}\right)+b\sum_i{\alpha_i y_i}=w^Tw+0=||w||^2$$