I want to know if it is possible to solve a QCQP problem with quadratic equality constraints in SDP. I know it is possible to convert a QCQP to an SDP by using the Shur complement. The following worked for me thus far:
$$
\begin{equation}
\begin{array}{cccccc}
\underset{x}{min} & x^{T}Q_{0}x+q_{0}^{T}x+c_{0} & & \underset{t,x}{min} & t\\
s.t & x^{T}Q_{i}x+q_{i}^{T}x+c_{i}\leq0 & \Longleftrightarrow & s.t & \left(\begin{array}{cc}
I & M_{0}x\\
x^{T}M_{0}^{T} & -c_{0}-q_{0}^{T}x+t
\end{array}\right) & \succeq0\\
& & & & \left(\begin{array}{cc}
I & M_{i}x\\
x^{T}M_{i}^{T} & -c_{i}-q_{i}^{T}x
\end{array}\right) & \succeq0\\
& i=1,2,…,m & & & i=1,2,…,m &
\end{array}
\end{equation}
$$
Where $M_{j}M_{j}^{T}=Q_{j}$ (Eigen Decomposition can also be used, thanks Alt)
However it seems this only applies to quadratic inequality constraints.
I considered converting the desired inequality constraints to equalities by using slack variables but I think that technique only works with linear constraints.
I also considered having two constraints like:
$$
\begin{equation}
\begin{array}{c}
x^{T}Qx+q^{T}x+c\leq0\\
-(x^{T}Qx+q^{T}x+c)\leq0
\end{array}
\end{equation}
$$
To force an equality but the second constraint is not convex so it wont work.
So is there another way to do it?
Edit: turns out quadratic equality constraints are non-convex thus cant be solved directly with this approach
Best Answer
Hint:
You don't need the constraints $M_{j}M_{j}^{T}=Q_{j}$. For example if $UDU^T$ is the eigen decomposition of $Q_j$, Then $M_j=UD^{\frac{1}{2}}$ (if $Q_j$ is symmetric)
Note that if you introduce new variable $M_j$, then the bilinear forms in the constraints make your problem non-convex, (Besides the fact that equality constraints with quadratic form $M_{j}M_{j}^{T}=Q_{j}$that you have to add also makes the problem non-convex.)