Transform semi-definite programming problem with linear matrix inequality(LMI) to SDP of standard form For a SDP problem with LMI, we can write it as:
$$minimize \quad c^Tx$$
$$ s.t. \quad x_1F_1+x_2F_2+...+x_nF_n+G \preceq 0$$
$$Ax=b$$
where $G,F_1,F_2,...,F_n \in S^k,A \in R^{p \times n}$.
Now I want to transform above form to SDP of standard form:
$$minimize \quad tr(CX)$$
$$s.t. \quad tr(A_iX)=b_i,i=1,2,...,p$$
$$X \succeq 0$$
Where $C,A_1,...,A_p \in S^n$.
Suppose $A=\begin{bmatrix} a^T_1 \\ a^T_2 \\ \vdots \\a^T_p\end{bmatrix}$,
I tried to diagonalize $c,x,a_i$ such that:
$$C=diag(c),X=diag(x),A_i=diag(a_i)$$
But I cannot figure out how satisfying constraint $X \succeq 0$. So what is the trick?
 A: You can dualize the problem but for that we need to assume that Slater's condition holds.
First, we build the Lagrangian
$$L(x,P,\mu):=c^Tx-\mathrm{trace}(PM(x))+\mu^T(Ax-b)$$
where $M(x):=x_1F_1+x_2F_2+...+x_nF_n+G$, $P$ is symmetric positive semidefinite and $\mu$ is a real vector.
The dual problem is, therefore,
$$\max_{P\succeq0,\mu\in\mathbb{R}^n}-(\mathrm{trace}(PG)-\mu^Tb)$$
such that
$$\mathrm{trace}(PF_i)+\mu^TAe_i=c_i.$$
where $\{e_1,\ldots,e_n\}$ is the natural basis of $\mathbb{R}^n$.
It is also possible to lump all the variables in the single matrix $\tilde P=\mathrm{diag}(P,U)$ where $U=\mathrm{diag}(\mu)$ to get
$$\max_{\tilde P}-\mathrm{trace}(\tilde PZ_0)$$
such that
$$\mathrm{trace}(\tilde PZ_i)=c_i$$
where $Z_0=\mathrm{diag}(G,\mathrm{diag}(b))$ and  $Z_i=\mathrm{diag}(F_i,\mathrm{diag}(Ae_i))$.
A: Instead of trying to rewrite it directly, since you already have a LMI it might be easier to write $Ax=b$ into the LMI.
Let
\begin{align*}
    &\min c^{\mathsf T} x \\
    &\text{s.t. } \sum_{i=1}^n x_i F_i - G \succcurlyeq 0,\ Ax=b
\end{align*}
be the given problem, where $G,\ F_i \in S^k$, $A \in \mathbb{R}^{p\times n}$, $x \in \mathbb{R}^n$, $b \in \mathbb{R}^p$.
Let $a_i$ be the columns of $A$, $i=1,\dots,n$.
We can rewrite $Ax=b$ as $\sum_{i=1}^n a_i x_i - b = 0$ element wise.
This can then be written as two constraints $\sum_{i=1}^n a_i x_i - b \ge 0$ and $\sum_{i=1}^n -a_i x_i - (-b) \ge 0$.
Now we can simply add this to the LMI, so we get $\sum_{i=1}^n x_i F'_i - G' \succcurlyeq 0$, for block matrices
\begin{align*}
    F'_i &= \begin{bmatrix} F_i & & \\ & \text{diag}(a_i) & \\ && -\text{diag}(a_i) \end{bmatrix} \\
    G' &= \begin{bmatrix} G && \\ & \text{diag}(b) & \\ &&-\text{diag}(b) \end{bmatrix}
\end{align*}
using that a block matrix is PSD if and only if its blocks are PSD, and a diagonal matrix is PSD if and only if its diagonal elements are nonnegative.
Now we are done, because the dual is
\begin{align*}
    &\max \langle G', X \rangle \\
    &\text{s.t. } \langle F'_i, X \rangle = c_i \quad i=1,\dots,n \\
    &\qquad X \succcurlyeq 0.
\end{align*}
