Is there an efficient method to solve the QCQP problem:
\begin{array}{ll} \text{minimize} & x^T Q x - f^Tx\\ \text{subject to} & x^TVx = 0\end{array}
where $Q$ is symmetric positive definite and $V$ is not necessary symmetric but should be semi-positive definite?
I saw this post but the constraint is equal to $1$ and I can't transform my problem in a similar way.
EDIT: Is there a way to solve the problem if $V$ is symmetric but it's not PSD?
Next is there an extension of the method for the following problem?: \begin{array}{ll} \text{minimize} & x^T Q x - f^Tx\\ \text{subject to} & x^TVx = 0 \\ \text{and subject to} & Cx=d\end{array}
EDIT: If $V$ is symmetric but it's not PSD I can find a subset of the solutions of $x^T V x = 0$ using svd. In fact I can define $V = U^T S W $ with $ S>0$ and diagonal, then $x^T V x = (U x)^T S (W x) $. A subset of solutions is the union of the null-space of $U$ and $W$. There are also other solutions that with this method I can't find. Have you any idea to improve the method?
If $V$ is PSD then so is $W=(V+V^T)/2$ and your constraint is equivalent to $x^TWx=0$. Now $W$ is symmetric PSD, so $x$ must be in the null-space of $W$ (or you take the Cholesky factorization $W=FF^T$ which turns your constraint into $F^Tx=0$). Therefore your constraint is in fact linear.