How can I form this second-order conic constraint?

Question

I am trying to show that if we have a real symmetric matrix $Q$ with one negative eigenvalue within a quadratic constraint:

$x^\top Qx+a^\top x +b \leq 0$

That this constraint can be formulated as a second-order conic constraint. My intial attempt at doing this was by performing an eigenvalue decomposition and then bringing the negative eigenvalue to the RHS in order to get an expression that is like:

$||x||_2\leq t$

Note that this was inspired by section 3.2.8 of https://docs.mosek.com/modeling-cookbook/cqo.html.

However I am unsure what to do about the additional $a^\top x +b$ terms here. Would we need to make certain assumptions about these terms?

Answer 1

You can use shift and rotation to rewrite the inequality as $$y^TBy+c\leq 0.$$ Or you can use the equality $${x}^T{Q}{x} + {a}^T{x} + b = \begin{pmatrix}x\\1\end{pmatrix}^T\begin{bmatrix}A&a/2\\a^T/2&b\end{bmatrix} \begin{pmatrix}x\\1\end{pmatrix}.$$

You can find related results searching fo "\(x^TAx+bx+c=0\)" on SearchOnMath.

Edit: Or you can write $${x}^T{Q}{x} + {a}^T{x} + b=(x-x_0)^TQ(x-x_0)+c.$$

It follows that $$-x_0^TQx-x^TQx_0=-2x_0^TQx=a^Tx,\qquad x_0^TQx_0+c=b.$$