Can an underdetermined system have a unique solution?

In my case, I am calling an underdetermined system as a system of linear equations where there are fewer equations than variables (unknowns). My textbook says the answer is false, however the internet says otherwise.

On the outside, it makes sense that an underdetermined system can't have a unique solution, but I am leaning towards trusting the internet on this one. Can someone explain how a unique solution exists? A proof might be helpful too, although hopefully not too complex.

Solution 1:

An underdetermined system is one with fewer equations than unknowns, so we can write this as a matrix equation $Ax = b$ with $A$ a matrix that has fewer rows than columns. This implies that solutions, if they exist, will not be unique. Two ways to see this:

Method 1. If you solve the equation $Ax = b$ by row reducing $A$ into echelon form, you will find that not every column in the echelon form can have a pivot. Therefore, when you write down the general solution, there will be free variables, leading to an infinite number of solutions.

Method 2. Since $A$ has fewer rows than columns, then $A$ is an $m \times n$ matrix with $m < n$. Then $\mathrm{rank}(A) \le m$, and since $\mathrm{rank}(A) + \mathrm{nullity}(A) = n$, then $\mathrm{nullity}(A) = n - \mathrm{rank}(A) \ge n - m > 0$. Therefore the null space of $A$ has dimension greater than $0$, so if $x_p$ is a particular solution to the equation, then any vector of the form $x_p + h$ is also a solution for any $h \in \mathrm{Nul}(A)$, and there are infinitely many choices for $h$.

Solution 2:

When solving $Ax=b$, with more unknown than equations, $A\in\mathbb{R}^{m\times n}$ with $m<n$.

Also $\operatorname{rank} A\leq m<n$. Since the linear system has a unique solution when $\operatorname{rank} A=n$, this can never occur.

As @Lazar commented, if there are additional conditions, you can get a unique solution. The linear equations don't provide the uniqueness though; the additional conditions do.