Problem: We consider a classification problem with two classes having equal probability and their corresponding points in R2 as follows: \begin{align*} &\text{Class } y_1 = 1: x_1 = (1,1)^T\\ &\text{Class } y_2 = -1: x_2 = (-1,-1)^T, x_3 = (1,0)^T, x_4 =(0,1)^T. \end{align*} Find the separating hyperplane given by SVM classifier.
My attempt: Consider the Lagragian function $$L(w,b,\alpha) = \dfrac{1}{2}\Vert w\Vert^2 + \sum_{i=1}^{4}\alpha_i(1-y_i(\langle w,x_i\rangle+b)),\ \alpha_i \ge 0.$$
KKT conditions for the primal problem: \begin{align*} &1-y_i(\langle w,x_i\rangle+b \le 0, \forall i= \overline{1,4} \tag{1}\\ &\alpha_i \ge 0, \forall i = \overline{1,4}. \tag{2}\\ &\alpha_i(1-y_i(\langle w,x_i\rangle+b) = 0 \tag{3}\\ &\dfrac{\partial L}{\partial w}(w,b,\alpha) = w - \sum_{i=1}^{4}\alpha_i y_i x_i = 0 \tag{4}\\ &\dfrac{\partial L}{\partial b}(w,b,\alpha) = - \sum_{i=1}^{4}\alpha_i y_i = 0 \tag{5} \end{align*}
Hence, the dual Lagrangian function is $$g(\alpha) = \dfrac{-1}{2}\alpha^T H\alpha + \alpha^T\cdot 1,$$ where $H = [H_{ij}]_{i,j=\overline{1,4}}$, with $H_{ij} = y_iy_j\langle x_i,x_j\rangle$. The dual problem now is \begin{align*} &\max_{\alpha} g(\alpha) \Leftrightarrow \min_{\alpha} \left\{\dfrac{1}{2}\alpha^T H \alpha - \alpha^T \cdot 1\right\}\\ \text{s.t }& \alpha_i \ge 0, \forall i = \overline{1,4}\\ &\sum_{i=1}^{4} \alpha_iy_i = 0. \end{align*}
KKT conditions for the dual problem: $$L_D(\alpha,\lambda) = \dfrac{1}{2}\alpha^T H \alpha - \alpha^T \cdot 1 + \lambda\sum_{i=1}^4 \alpha_i y_i$$ \begin{align*} &\alpha_i \ge 0 \tag{6}\\ &\sum_{i=1}^4 \alpha_iy_i = 0 \tag{7}\\ &H\alpha -1 + \lambda y = 0 \tag{8} \end{align*} Combining (7) and (8), we have the linear systems of 5 equations and 5 variables. I expected that this system has a solution but it did not. So, I wonder why it happened because if the dual problem has no solution then the primal does. But by plotting these data points, we can easily see the separating hyperplane.