There are a few questions on this topic already. However, none of them really answer my question. The most relevant are these:

I have a quadratic problem with quadratic constraints, and my constraints are equalities: $$\text{mimize}\quad x^TQ_0x+q_0^Tx\\ \text{s. t.}\quad x^TQ_ix+q_i^Tx=0$$ which can be rewritten as: $$\text{mimize}\quad x^TQ_0x+q_0^Tx\\ \text{s. t.}\quad x^TQ_ix+q_i^Tx\le 0\\ \quad\quad\quad\quad x^T(-Q_i)x-q_i^Tx\le 0$$ On the literature I've read, the only restriction for the problem to be convex are that the matrices $Q$ have to be positive semi-definite, which is satisfied (in my case) for both restrictions.

Does the equality make the problem nonconvex and can someone give me some references about this? Or, since the semi-definiteness is preserved, is my problem is still convex?

Lastlty, if it is nonconvex, does going to a SOCP or SDP help me?

Thank you

• As we know semi-definiteness conclude convexity, so your problem is remained convex. – Nosrati Jan 6 '16 at 16:12
• Thank you very much for your comment, so are you saying that the problem is convex if and only if Q is semi-definite? Or in other words the that it is sufficient to guarantee convexity? – strangelyput Jan 7 '16 at 10:35
• Your second constraint is $x^T(-Q_i)x - q_i^Tx\leq 0$, hence the matrix in the quadratic constraint is negative semidefinite if $Q_i$ is positive semidefinite. For $x^TAx+b^Tx \leq 0$ to be a convex set, $A$ has to be positive semidefinite. You cannot just disregard the signs – Johan Löfberg Jan 7 '16 at 11:58
• Johan, please check the comment to your answer: math.stackexchange.com/questions/1602137/… – strangelyput Jan 7 '16 at 13:08

Quadratic equalities are never convex. Simply consider the trivial scalar case $x^2=1$, which has two distinct feasible points $-1$ and $1$.

As it is nonconvex, you cannot convert it to an SOCP, so your final question doesn't really make any sense.

• that might be true but in that case the semi-definiteness is not preserved, so that is not really my question – strangelyput Jan 7 '16 at 10:28
• How do you mean semi-definiteness ever could be preserved when writing a quadratic equality as two inequalities. A quadratic constraint is never convex. No equalities beyond affine are convex. – Johan Löfberg Jan 7 '16 at 11:54
• To the best of my knowledge, for a matrix to be positive semi-definite its eigenvalues must all be non-negative. So for some particular cases of matrices, both $Q_i$ and $-Q_i$ can be both positive semi-definite. For example, [0 1; 0 0] and [0 -1; 0 0]. – strangelyput Jan 7 '16 at 12:09
• Convexity is checked on the hessian, which will be $(Q_i + Q_i^T)/2$, i.e., the $Q_i$ matrix has to be symmetric. If you check eigenvalues on those symmetrized matrices, you will see they are indefinite. – Johan Löfberg Jan 7 '16 at 14:29
• Ok, that makes sense, because on the example $Q_i$ is not Hermitian.Its Hermitian version [0 1; 1 0] is indefinite. I was not aware of this requirement. Thank you for your help. Do you see any hope for this problem to be written in a convex framework? – strangelyput Jan 7 '16 at 15:16

For the constraints, we normally require that the feasible set be a convex set, while the objective function should be convex. In general, the set of points (or vectors) satisfying a quadratic equality constraint may not be a convex set.

For example, take the scalar case where Q = 1 (positive definite) and the quadratic equation is $$\begin{equation*} x^T(1)x = 9 \\ \Rightarrow x^2 = 9 \end{equation*}$$ The feasible set, in this case, is {-3,3}. The line joining the elements -3 and 3 does not satisfy the quadratic equality. In particular, taking $$\alpha$$ as 0.5, the square of zero is clearly not 9. Hence, despite Q being positive definite, the constraint is not convex.

See this answer: Michael Grant (https://math.stackexchange.com/users/52878/michael-grant), Why is the constraint $$\|w\| = 1$$ non-convex?, URL (version: 2015-05-28): https://math.stackexchange.com/q/1301592.