The problem to solve
Using our definition of the kernel We can rewrite again the Lagrangian
{\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{z}_j,
subject to the constraints
\lambda_i\geq 0 ,
\sum_i\lambda_iy_i=0 in terms of a convex optimization problem
\frac{1}{2} \boldsymbol{\lambda}^T\begin{bmatrix} y_1y_1K(\boldsymbol{x}_1,\boldsymbol{x}_1) & y_1y_2K(\boldsymbol{x}_1,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_1,\boldsymbol{x}_n) \\
y_2y_1K(\boldsymbol{x}_2,\boldsymbol{x}_1) & y_2y_2(\boldsymbol{x}_2,\boldsymbol{x}_2) & \dots & \dots & y_1y_nK(\boldsymbol{x}_2,\boldsymbol{x}_n) \\
\dots & \dots & \dots & \dots & \dots \\
\dots & \dots & \dots & \dots & \dots \\
y_ny_1K(\boldsymbol{x}_n,\boldsymbol{x}_1) & y_ny_2K(\boldsymbol{x}_n\boldsymbol{x}_2) & \dots & \dots & y_ny_nK(\boldsymbol{x}_n,\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] .
If we add the slack constants this leads to the additional constraint
0\leq \lambda_i \leq C .
We can rewrite this (see the solutions below) in terms of a convex optimization problem of the type
\begin{align*}
&\mathrm{min}_{\lambda}\hspace{0.2cm} \frac{1}{2}\boldsymbol{\lambda}^T\boldsymbol{P}\boldsymbol{\lambda}+\boldsymbol{q}^T\boldsymbol{\lambda},\\ \nonumber
&\mathrm{subject\hspace{0.1cm}to} \hspace{0.2cm} \boldsymbol{G}\boldsymbol{\lambda} \preceq \boldsymbol{h} \hspace{0.2cm} \wedge \boldsymbol{A}\boldsymbol{\lambda}=f.
\end{align*}
Below we discuss how to solve these equations. Here we note that the matrix \boldsymbol{P} has matrix elements p_{ij}=y_iy_jK(\boldsymbol{x}_i,\boldsymbol{x}_j) .
Given a kernel K and the targets y_i this matrix is easy to set up. The constraint \boldsymbol{y}^T\boldsymbol{\lambda}=0 leads to f=0 and \boldsymbol{A}=\boldsymbol{y} . How to set up the matrix \boldsymbol{G} is discussed later. Here note that the inequalities 0\leq \lambda_i \leq C can be split up into
0\leq \lambda_i and \lambda_i \leq C . These two inequalities define then the matrix \boldsymbol{G} and the vector \boldsymbol{h} .