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Can we formulate as a linear matrix inequality (LMI) one of the regions in which a concave quadratic polynomial is negative (without solving its roots)?

Let the quadratic polynomial $f \in {\Bbb R} [x]$ be concave and with positive discriminant (thereby ensuring $2$ real roots). I would like to express the region $f(x) \le 0$ intersected with $x \le a$ where $a$ is some number between the roots of $f(x)$. For instance, $f(x) = -x^2 + 2x + 1$ has two roots, $x_0 := 1-\sqrt{2}$ and $x_1 := 1+\sqrt{2}$. I want to express as an LMI the region $x \le x_0$ but without solving explicitly the roots of $f(x)$.

Rodrigo de Azevedo
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Morad
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  • $\def\ed{\stackrel{\text{def}}{=}}$ I presume you mean positive discriminant. If the quadratic is $\ -x^2+bx+c\ $ with positive discriminant $\ b^2+4c\ ,$ then $\ x=\frac{b}{2}\ $ always lies between the quadratic's two roots. Therefore, the linear matrix inequality $$A_0+x_1A_1\succeq0\ ,$$ with $\ A_0\ed\left[\frac{b}{2}\right]\ $ and $\ A_0\ed[-1]\ $ would appear to satisfy your criterion. But I'm a little puzzled as to why you'd want to write a simple inequality,$\ x\le\frac{b}{2}\ ,$ as a linear matrix inequality with $\ 1\times1\ $ matrices. – lonza leggiera Feb 28 '25 at 09:14
  • Frankly, I still do not understand why you need an LMI. Wouldn't introducing the constraint x <= 1 - sqrt(2) be much simpler? – Rodrigo de Azevedo Mar 05 '25 at 12:20
  • This is part of much bigger optimization and the root cannot be found directly. I am formulating now a similar question on the matrix case, in hope to make a progress towards the most general case. – Morad Mar 05 '25 at 12:55
  • https://math.stackexchange.com/questions/5042346/kalman-filter-solution-via-lmi @RodrigodeAzevedo – Morad Mar 05 '25 at 13:28

2 Answers2

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$$ q:= x^2 - 2 x - 1 = (x - 1)^2 - 2 = (x - 1)^2 - \left( \sqrt{2} \right)^2 = (1 - x)^2 - \left( \sqrt{2} \right)^2 \geq 0 $$

The semialgebraic set (and spectrahedron) $\left] - \infty, 1 - \sqrt{2} \right]$ can be defined by the following linear matrix inequality (LMI)

$$ \begin{bmatrix} 1 - x & \sqrt{2} \\ \sqrt{2} & 1 - x \end{bmatrix} \succeq {\bf O}_2 $$

which, using Sylvester's criterion, encapsulates the linear inequality $1 - x \geq 0$ (or, $x \leq 1$) twice and the quadratic inequality $q = (1 - x)^2 - \left( \sqrt{2} \right)^2 \geq 0$ once. Thus, the root $1 - \sqrt{2}$ can be found via the following semidefinite program (SDP)

$$ \boxed{\begin{array}{ll} \underset {x \in {\Bbb R}} {\text{maximize}} & x \\ \text{subject to} & \begin{bmatrix} 1 - x & \sqrt{2} \\ \sqrt{2} & 1 - x \end{bmatrix} \succeq {\bf O}_2 \end{array}} $$

whereas the root $1 + \sqrt{2}$ can be found via the following SDP

$$ \begin{array}{ll} \underset {x \in {\Bbb R}} {\text{minimize}} & x \\ \text{subject to} & \begin{bmatrix} x - 1 & \sqrt{2} \\ \sqrt{2} & x - 1 \end{bmatrix} \succeq {\bf O}_2 \end{array} $$

whose feasible region is defined by an LMI that encapsulates the linear inequality $x - 1 \geq 0$ (or, $x \geq 1$) twice.

Numerical experiment

In Python + NumPy + CVXPY:

>>> import numpy as np
>>> from cvxpy import *
>>> I_2 = np.identity(2)
>>> H_2 = np.ones((2,2)) - I_2
>>> x = Variable()
>>> SDP = Problem(Maximize(x), [ ((1-x) * I_2) + (np.sqrt(2) * H_2) >> 0 ])
>>> SDP.solve()
-0.4142140469344502

How close is this maximum to the root $1 - \sqrt{2}$?

>>> 1 - np.sqrt(2)
-0.41421356237309515
>>> x.value - (1 - np.sqrt(2))
-4.845613550408245e-07
Rodrigo de Azevedo
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  • my suggestion was to include it in your solution since my question was on general quadratic. It did not help the general case, I consider posting it as a new question. – Morad Mar 04 '25 at 06:44
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    @Morad Earlier today, I worked on the generalization $x^2 - 2 b x + c \geq 0$, completed the square and eventually arrived at an LMI that defines the spectrahedron $\left] -\infty, b - \sqrt{b^2 - c} \right]$. This general case is dealt with in a manner very similar to what I wrote in my answer. Why don't you copy+paste my answer, edit it to handle the general case and post it as an answer? It should be only a few minutes of work. – Rodrigo de Azevedo Mar 04 '25 at 12:09
  • Sure, I added. Does it work for quadratic polynomials of (psd) matrices intersected with $\mathtt{Tr}(XW)\le \alpha$? for some $W\succeq0$ and $\alpha\in\mathbb{R}$? – Morad Mar 04 '25 at 19:21
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The general case is $f(x) = ax^2 + bx + c$ with $a>0$, so w.l.o.g. we can characterize the intersection $g(x) = x^2 - 2 bx + c > 0$ with $x < \alpha$ or with $x > \alpha$, where $\alpha$ falls between the two real roots of $g$. It is assumed that the discriminant is positive, i.e., $b^2-c > 0$, to ensure that there are two distinct real roots.

Solution: use the completion of the squares to write $$g(x) = (x-b)^2 - \left(\sqrt{b^2 - c}\right)^2$$ where the square root exists due to the positive discriminant. We can now solve both scenarios using the Schur complement trick:

\begin{align} \{x > b\} \bigcap \left\{(x-b)^2 - \left(\sqrt{b^2 - c}\right)^2>0 \right\} \iff \begin{pmatrix} x-b & \sqrt{b^2-c} \\ \sqrt{b^2-c} & x-b \end{pmatrix} \succ 0. \end{align}

\begin{align} \{x < b\} \bigcap \left\{ (x-b)^2 - \left(\sqrt{b^2 - c}\right)^2>0 \right\} \iff \begin{pmatrix} b-x & \sqrt{b^2-c} \\ \sqrt{b^2-c} & b-x \end{pmatrix} \succ 0. \end{align}

Rodrigo de Azevedo
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Morad
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