When does an under-specified linear sy stem have a unique solution?

by Erel Segal-Halevi   Last Updated January 11, 2019 12:20 PM

Consider the system of two equations with three real variables $x,y,z$: $$ a_{11}\cdot x + a_{12} \cdot y + a_{13} \cdot z = b_1 \\ a_{21} \cdot (1-x) + a_{22} \cdot (1-y) + a_{23} \cdot (1-z) = b_2 $$ where $x,y,z\geq 0$.

Since there are less equations than variables, this system may have infinitely many solutions. My question is: is there a simple condition on $b_1,b_2$ (as a function of the coefficients $a_{ij}$), that guarantees that this system has a unique solution?

Answers 2

For example if you have that $a_{11},a_{12},a_{13}>0$ and $b_1=0$ your only solution for the first equation is $x=y=z=0$ that is a solution for the second equation if and only if


So in the case in which \begin{cases} b_1=0\\ b_2=a_{21}+a_{22}+a_{23} \end{cases}

your only solution is $x=y=z=0$

The geometric idea is no difficult. Your system of equations represent the locus of the intersection of two planes related by your equations but in general in $\mathbb{R}^3$ the intersection of two planes is the empty set or a plane (if the two planes are equal) or a line. There is not the case in which the intersection is one single point. But In your case you must consider also the condition $x,y,z\geq 0$ so you can consider $b_1,b_2$ such that the two different planes intersect them in a line and that line passes on the origin and its intersection with the locus $x,y,z\geq 0$ is only the origin. In this case your only solution will be the origin.

Federico Fallucca
Federico Fallucca
January 11, 2019 11:24 AM

In general the answer is no: you are looking for a solution $x\geq0$ to the problem $Ax=b$. Suppose there is no vector $y$ (of right size) such that $A^Ty>0$ (1). Then by the dual version of Gordan's lemma there exists some $z>0$ such that $Az=0$. Therefore in this case, either there is no solution $x\geq0$, or $x\geq0$ produces $Ax=b$, and then $A(x+z)=b$ with $x+z\geq0$, so you have at least two solutions. Since your $A$ is completely arbitrary, you have cases satisfying (1).

Jose Brox
Jose Brox
January 11, 2019 11:29 AM

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