How can I minimize a quadratic function which some of the quadratic terms have a coefficient of zero?

e.g. $\min x_1^2 + x_1 + x_2$

(subject to some linear constraints on $x_i$)

As a quadratic programming problem, this is

$\min q^\intercal x + \frac{1}{2}x^\intercal Q x$


$Q = \begin{pmatrix}2 & 0 \\0 & 0\end{pmatrix}, q = \begin{pmatrix}1 \\ 1 \end{pmatrix}$

However $Q$ is not positive-definite and therefore a standard QP solver cannot be used. What other methods can I use?

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    $\begingroup$ What are the linear constrains? $\endgroup$ – Vasily Mitch Oct 8 '19 at 8:55
  • $\begingroup$ I wrote this example without constraints thinking they do not matter for the question, but perhaps I'm mistaken-would they? $\endgroup$ – njd42 Oct 8 '19 at 9:18
  • $\begingroup$ Yes, they do. Depending on the constrains, the problem can have on have not a reasonable answer $\endgroup$ – Vasily Mitch Oct 8 '19 at 9:20
  • $\begingroup$ $x_1 - x_2 \le 0$ and $-x_1 - x_2 \le 0$ and $|x_1| < c$ for some $c>0$ $\endgroup$ – njd42 Oct 8 '19 at 9:25
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    $\begingroup$ A standard QP solver will handle a semidefinite $Q$. In practice it is very rare that all optimiziation variables appear in the quadratic term. $\endgroup$ – Michal Adamaszek Oct 8 '19 at 10:54

Matrix $Q$ has zero eigenvector that isn't orthogonal to $q$. That means that without constraints the function has no global minimum.

So the minimum of the problem is achieved on a boundary. We have 4 linear boundaries and 3 points (vertices). We solve boundary problems with Lagrange multipliers.

For boundaries $x_1=\pm c$ there is no minimum. For boundary $x_1-x_2=0$, there is minimum $f=-1$ at $(-1,-1)$, which lies outside of condition $x_1+x_2\ge0$. For boundary $x_1+x_2=0$ there is minimum $f=0$ at $(0,0)$ which coincides with one vertex. For other two vertices $(-c,c)$ and $(c,c)$, $f=c^2>0$ and $f=c^2+2c>0$.

Thus, we conclude that the minimum $f=0$ is achieved at vertex $(0,0)$.


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