Obtained from "Engineering Design via Surrogate Modelling: A Practical Guide, Forrester", The convex problem is written as
Maximize,
$\frac{1}{2}\sum_{i,j=1}^{n} (\alpha^{+(i)} - \alpha^{-(i)}) (\alpha^{+(j)} - \alpha^{-(j)})\Psi(x^i,x^j) - \epsilon\frac{1}{2}\sum_{i=1}^{n}(\alpha^{+(i)} - \alpha^{-(i)}) + \sum_{i=1}^{n}y^i(\alpha^{+(i)} - \alpha^{-(i)}) $
Subject to,
$ \sum_{i=1}^{n}(\alpha^{+(i)} - \alpha^{-(i)}) = 0 $ and $\alpha^{\pm}\in[0,C/n] $
This in turn is changed into a minimize problem by just multiplying by -1, Thus the Matrix formulation is as written.
Minimize, \begin{equation} \frac{1}{2} \begin{bmatrix} \alpha^+\\-\alpha^-\\ \end{bmatrix}^T \begin{bmatrix} \Psi & -\Psi\\ -\Psi & \Psi \end{bmatrix} \begin{bmatrix} \alpha^+\\-\alpha^-\\ \end{bmatrix} + \begin{bmatrix} 1^T\epsilon - y\\ 1^T\epsilon + y \end{bmatrix}^T \begin{bmatrix} \alpha^+\\-\alpha^-\\ \end{bmatrix} \end{equation}
Subject to, \begin{equation} 1^T\begin{bmatrix} \alpha^+\\-\alpha^-\\\end{bmatrix} = 0 \end{equation} \begin{equation} \alpha^+,\alpha^-\in [0, C/n] \end{equation}
I would like to know the steps for deriving the matrix formulation from the summation.
There isn't really a step-by-step method for getting this kind of compact (i.e. block-matrix based) formula. However, here is a method for writing the a quadratic expression in terms of matrices.
Suppose we have a sum of the form $$ S = \sum_{i \leq j}^N a_{ij} v^iv^j + \sum_{i=1}^N b_i v^i. $$ Note that if we take $v = (\alpha^{+(1)}, \dots , \alpha^{+(N)},\alpha^{-(1)}, \dots , \alpha^{-(N)})$, then expanding and rearranging the products in the expression that you are meant to minimize will give you something of the above form. We can write $$ S = v^TMv + b^Tv, $$ where $b = (b_1,\dots,b_N)^T$ and $$ M_{ij} = \begin{cases} a_{ij} & i=j\\ a_{ij}/2 & i \neq j. \end{cases} $$ To understand how your particular objective function was rewritten, I suggest that you apply this idea for the cases of $n=2,n=3$. After that, I think it should be clear why the given matrix formulas work.