Solving the above problem, yields the values of \( \lambda_i \). To find the coefficients of your hyperplane we need simply to compute
$$ \boldsymbol{w}=\sum_{i} \lambda_iy_i\boldsymbol{x}_i. $$With our vector \( \boldsymbol{w} \) we can in turn find the value of the intercept \( b \) (here in two dimensions) via
$$ y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1, $$resulting in
$$ b = \frac{1}{y_i}-\boldsymbol{w}^T\boldsymbol{x}_i, $$or if we write it out in terms of the support vectors only, with \( N_s \) being their number, we have
$$ b = \frac{1}{N_s}\sum_{j\in N_s}\left(y_j-\sum_{i=1}^n\lambda_iy_i\boldsymbol{x}_i^T\boldsymbol{x}_j\right). $$With our hyperplane coefficients we can use our classifier to assign any observation by simply using
$$ y_i = \mathrm{sign}(\boldsymbol{w}^T\boldsymbol{x}_i+b). $$Below we discuss how to find the optimal values of \( \lambda_i \). Before we proceed however, we discuss now the so-called soft classifier.