I have the following SDP problem:
max: $Tr(CX)$
subject to: $X \geq 0, I - X \geq 0$.
I want to convert it into the standard form specified by CSDP (I'm using the callable C interface), which is:
max: $Tr(CX)$
subject to: $\forall_i, Tr(A_iX) = a_i, X \geq 0$.
The SDP is part of an algorithm which iteratively increases the dimensions of $X$ until the value of $\max\ Tr(CX)$ converges.
Is there any efficient way? $X$ is already known to be sparse, if that helps.
Suppose that $X \in \mathbb{R}^{n \times n}$, and $X$ is positive semidefinite. For convenience, define
\begin{align} \langle A, B\rangle = \sum_{k, l} A_{kl}B_{kl} \end{align}
for square matrices $A, B$ of equal size; this operation corresponds to the trace of the matrix product.
So your formulation currently looks like
\begin{align} \max \langle{C, X}\rangle \\ \mathrm{s.t.}\,\, X \succeq 0,\,\,I - X \succeq 0, \end{align}
where $X \succeq 0$ means that $X$ is positive semidefinite.
Essentially, I'd create a positive semidefinite matrix of slack variables $S \in \mathbb{R}^{n \times n}$ such that $S + X = I$, and $S \succeq 0$. Let $E_{ij}$ is a matrix of consisting of all zeros except for the $(i,j)$th element, which is set to 1. Also, define the matrices
\begin{align} X' = \left[\begin{array}{cc}X & 0 \\ 0 & S\end{array}\right], \\ C' = \left[\begin{array}{cc}C & 0 \\ 0 & 0\end{array}\right], \\ A_{ij} = \left[\begin{array}{cc} E_{ij} & 0 \\ 0 & E_{ij}\end{array}\right]. \end{align}
Then, (I think,) the following reformulation works:
\begin{align} \max \langle{C', X'}\rangle \\ \mathrm{s.t.}\,\, X' \succeq 0,\,\,\langle A_{ij}, X' \rangle = \delta_{ij},\,\,i,j=1,\ldots,n \end{align}
where $\delta_{ij}$ is the Kronecker delta symbol.
