I have the following QCQP:
$$ \underset{\mathbf{x}}{minimize} \quad \mathbf{x}^{T} H \mathbf{x} + \mathbf{p}^{T} \mathbf{x}$$ $$ subject \; to \quad \mathbf{x}^{T}Q_{i}\mathbf{x} + \mathbf{q}_{i}^{T}\mathbf{x} + r_{i} \leq 0, \quad i = 1, 2, \ldots, p $$
Where $\mathbf{x} \in \mathbb{R}^{n}$. I convert it into an SDP as follows. The first step is to rewrite in epigraph form:
$$ {minimize} \quad \delta $$ $$ subject \; to \quad \mathbf{x}^{T} H \mathbf{x} + \mathbf{p}^{T} \mathbf{x} \leq \delta$$ $$ \quad \mathbf{x}^{T}Q_{i}\mathbf{x} + \mathbf{q}_{i}^{T}\mathbf{x} + r_{i} \leq 0, \quad i = 1, 2, \ldots, p$$
Since $H$ is PSD, we can take its Cholesky factorization, giving us $H = \tilde{H}^{T}\tilde{H}$. Defining $\mathbf{\tilde{x}} = [\delta, \mathbf{x}]^{T}$ The first constraint can therefore be written as:
$$ G(\mathbf{\tilde{x}}) = \begin{bmatrix} I_{n} & \tilde{H} \mathbf{x} \\ (\tilde{H} \mathbf{x})^{T} & \delta - \mathbf{p}^{T}\mathbf{x} \end{bmatrix} \geq 0 $$
Similarly, $Q_{i} = \tilde{Q_{i}}^{T}\tilde{Q_{i}}$ the other equations can be written as: $$ F_{i}(\mathbf{\tilde{x}}) = \begin{bmatrix} I_{n} & \tilde{Q_{i}} \mathbf{x} \\ (\tilde{Q_{i}} \mathbf{x})^{T} & -\mathbf{q_{i}}^{T}\mathbf{x} - r_{i} \end{bmatrix} \geq 0 $$
So we can put all of these together, and write:
$$ E(\mathbf{\tilde{x}}) = diag\{ G(\mathbf{\tilde{x}}), F_{1}(\mathbf{\tilde{x}}), \ldots, F_{p}(\mathbf{\tilde{x}})\} \geq 0 $$
So we get the SDP problem:
If we let $\mathbf{\tilde{c}} =[1, 0, 0, \ldots, 0]^{T}$ then clearly, $\delta = \mathbf{\tilde{c}}^{T} \mathbf{\tilde{x}}$
$$ \underset{\mathbf{\tilde{x}}}{minimize} \quad \mathbf{\tilde{c}}^{T} \mathbf{\tilde{x}}$$ $$ subject \; to \quad E(\mathbf{\tilde{x}}) \geq 0$$
What I am stuck with is how to convert that into something of the form:
$$ \underset{X}{minimize} \quad trace(CX)$$ $$ subject \; to \quad trace(A_{i} X) = b_{i} \quad i = 1, 2, \ldots, p$$ $$ X \geq 0$$
How can I do this? Is it even possible?
My attempt at a solution, following the suggestions in the comments:
The first step is to convert all the LMI matrices into LMIs proper. To do so, we notice that we can convert an LMI matrix:
\begin{equation} \begin{bmatrix} I_{n} & A \mathbf{x} + \mathbf{b} \\ (A \mathbf{x} + \mathbf{b})^{T} & \mathbf{c}^{T}\mathbf{x} + d \end{bmatrix} \geq 0 \end{equation}
Into the LMI equation $F_{0} + \displaystyle\sum\limits_{j=1}^{N} x_{j} F_{j} \geq 0$ by:
\begin{equation} F_{0} = \begin{bmatrix} I_{n} & \mathbf{b} \\ \mathbf{b}^{T} & d \end{bmatrix} \end{equation}
\begin{equation} F_{j} = \begin{bmatrix} \mathbf{0}_{n} & \mathbf{a}_{j} \\ \mathbf{a}_{j}^{T} & c_{j} \end{bmatrix} \end{equation}
Where $\mathbf{a}_{j}$ is the $j$th column of $A$, and $c_{j}$ is the $j$th element of $\mathbf{c}$.
So lets apply that to each LMI matrix. First up is $G(\mathbf{\tilde{x}}) \geq 0$. This gives us:
\begin{equation} G_{0} = \begin{bmatrix} I_{n} & \mathbf{0}_{n} \\ \mathbf{0}_{n}^{T} & \delta \end{bmatrix} \end{equation}
\begin{equation} G_{j} = \begin{bmatrix} \mathbf{0}_{n} & \mathbf{\tilde{h}}_{j} \\ \mathbf{\tilde{h}}_{j}^{T} & -p_{j} \end{bmatrix} \end{equation}
Where $\mathbf{\tilde{h}}_{j}$ is the $j$th column of $\tilde{H}$, and $p_{j}$ is the $j$th element of $\mathbf{p}$. Therefore, for $G(\mathbf{\tilde{x}}) \geq 0$, we have $G_{0} + \displaystyle\sum\limits_{j=1}^{N} x_{j} G_{j} \geq 0$.
Next up are the $F_{i}(\mathbf{\tilde{x}}) \geq 0$. We have now:
\begin{equation} F_{i_{0}} = \begin{bmatrix} I_{n} & \mathbf{0}_{n} \\ \mathbf{0}_{n}^{T} & -r_{i} \end{bmatrix} \end{equation}
\begin{equation} F_{i_{j}} = \begin{bmatrix} \mathbf{0}_{n} & \mathbf{\tilde{q}}_{i_{j}} \\ \mathbf{\tilde{q}}_{i_{j}}^{T} & q_{i_{j}} \end{bmatrix} \end{equation}
Where $\mathbf{\tilde{q}}_{i_{j}}$ is the $j$th column of $\tilde{Q}_{i}$, and $q_{i_{j}}$ is the $j$th element of $\mathbf{q}_{i}$. Therefore, for $F_{i}(\mathbf{\tilde{x}}) \geq 0$, we have $F_{i_{0}} + \displaystyle\sum\limits_{j=1}^{N} x_{j} F_{i_{j}} \geq 0$. So we put this together, to get:
\begin{align} minimize& \quad \mathbf{\tilde{c}}^{T} \mathbf{\tilde{x}} \nonumber \\ subject \; to& \quad G_{0} + \displaystyle\sum\limits_{j=1}^{N} x_{j} G_{j} \geq 0 \nonumber \\ & \quad F_{i_{0}} + \displaystyle\sum\limits_{j=1}^{N} x_{j} F_{i_{j}} \geq 0 \quad i=1,2,\ldots p \nonumber \\ \end{align}
Now the dual of this function can be obtained by letting:
\begin{align} \mathbf{\tilde{y}} &= -\mathbf{\tilde{x}} \nonumber \\ \mathbf{\tilde{b}} &= \mathbf{\tilde{c}} \nonumber \\ C_{0} &= G_{0} \nonumber \\ A_{j} &= G_{j} \nonumber \\ C_{i_{0}} &= F_{i_{0}} \nonumber \\ A_{i_{j}} &= F_{i_{j}} \nonumber \\ \end{align}
Which gives us:
\begin{align} maximize & \quad \mathbf{\tilde{b}}^{T} \mathbf{\tilde{y}} \nonumber \\ subject \; to& \quad C_{0} - \displaystyle\sum\limits_{j=1}^{N} y_{j} A_{j} \geq 0 \nonumber \\ & \quad C_{i_{0}} - \displaystyle\sum\limits_{j=1}^{N} y_{j} A_{i_{j}} \geq 0 \quad i=1,2,\ldots p \nonumber \\ \end{align}
Adding a slack matrix gives us:
\begin{align} maximize & \quad \mathbf{\tilde{b}}^{T} \mathbf{\tilde{y}} \nonumber \\ subject \; to& \quad \displaystyle\sum\limits_{j=1}^{N} y_{j} A_{j} + S_{0} = C_{0} \nonumber \\ & \quad \displaystyle\sum\limits_{j=1}^{N} y_{j} A_{i_{j}} + S_{i_{0}} = C_{i_{0}} \quad i=1,2,\ldots p \nonumber \\ \end{align}
And this is where I am currently getting lost. How can I construct my matrix $X$? Is the matrix $C$ equal to $C_{0}$? Also, how do I fit in my $b_{i}$ in the primal? Is my derivation even correct so far?
Following Johnan Loefburg's article 'Dualize it: software for automatic primal and dual conversions of conic programs' I get:
$$A(X) = [trace(A_{1}X), trace(A_{2}X), \ldots, trace(A_{N}X), trace(A_{1_{1}} X), trace(A_{1_{2}}X), \ldots, trace(A_{1_{N}}X), trace(A_{2_{1}}X), \ldots, trace(A_{p_{1}}X), \ldots, trace(A_{p_{N}}X)]^{T} $$ $$C = C_{0} + \displaystyle\sum\limits_{i=1}^{N} C_{i_{0}} $$ $$b = [\delta \mathbb{1}_{N}^{T}, \tilde{b}_{1}\mathbb{1}_{N}^{T}, \ldots, \tilde{b}_{N}\mathbb{1}_{N}^{T}]^{T} $$
I get:
$$ minimize \quad trace(CX) $$ $$ subject \; to \quad A(X) = b$$ $$ X \geq 0$$
What do you think? Is this correct?