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

Quadratic optimisation with quadratic equality constraints

Quadratic Equality Constraints via SDP

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

  • $\begingroup$ As we know semi-definiteness conclude convexity, so your problem is remained convex. $\endgroup$ – Nosrati Jan 6 '16 at 16:12
  • $\begingroup$ 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? $\endgroup$ – strangelyput Jan 7 '16 at 10:35
  • $\begingroup$ 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 $\endgroup$ – Johan Löfberg Jan 7 '16 at 11:58
  • $\begingroup$ Johan, please check the comment to your answer: math.stackexchange.com/questions/1602137/… $\endgroup$ – 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.

  • $\begingroup$ that might be true but in that case the semi-definiteness is not preserved, so that is not really my question $\endgroup$ – strangelyput Jan 7 '16 at 10:28
  • $\begingroup$ 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. $\endgroup$ – Johan Löfberg Jan 7 '16 at 11:54
  • $\begingroup$ 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]. $\endgroup$ – strangelyput Jan 7 '16 at 12:09
  • $\begingroup$ 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. $\endgroup$ – Johan Löfberg Jan 7 '16 at 14:29
  • $\begingroup$ 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? $\endgroup$ – strangelyput Jan 7 '16 at 15:16

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