We can rewrite {\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, and its constraints in terms of a matrix-vector problem where we minimize w.r.t. \lambda the following problem \frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1\boldsymbol{x}_1^T\boldsymbol{x}_1 & y_1y_2\boldsymbol{x}_1^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_1^T\boldsymbol{x}_n \\ y_2y_1\boldsymbol{x}_2^T\boldsymbol{x}_1 & y_2y_2\boldsymbol{x}_2^T\boldsymbol{x}_2 & \dots & \dots & y_1y_n\boldsymbol{x}_2^T\boldsymbol{x}_n \\ \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots \\ y_ny_1\boldsymbol{x}_n^T\boldsymbol{x}_1 & y_ny_2\boldsymbol{x}_n^T\boldsymbol{x}_2 & \dots & \dots & y_ny_n\boldsymbol{x}_n^T\boldsymbol{x}_n \\ \end{bmatrix}\boldsymbol{\lambda}-\mathbb{1}\boldsymbol{\lambda}, subject to \boldsymbol{y}^T\boldsymbol{\lambda}=0 . Here we defined the vectors \boldsymbol{\lambda} =[\lambda_1,\lambda_2,\dots,\lambda_n] and \boldsymbol{y}=[y_1,y_2,\dots,y_n] .