Least-squares QCQP problem with quadratic equality constraint equals to zero

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.
• Thanks a lot, I was arrived very closed to this idea but I didn't find a way to compute choleky factorization for PSD matrices. Anyway if I know the null-space of $W$ I solved but I don't understand why the null-space of $W$ is equivalent to the null-space of $F^T$ Commented Mar 29, 2022 at 12:08
• @Andrea993 $0 = x^TWx = x^T(FF^T)x = (F^Tx)^T(F^Tx) = \|F^Tx\|_2^2$ so $F^Tx=0$. Commented Mar 29, 2022 at 12:11