Can we solve an algebraic system where the number of equations is less than the number of unknowns?

Question

Is it necessary to always have the number of equations >= number of unknown variables to solve the problem? Can we have the question where this is not true?

Answer 1

It depends what you mean by solve. You can describe the solution set of the system of (linear) equations regardless of how many there are, but they may not determine a unique solution. To get a single solution you must have at least as many equations as variables. Think about determining a point in the plane with lines, you need at least two. Similarly in $\mathbb{R}^3$, three planes are needed.

Answer 2

Technically, no.

Consider $x^2 + y^2 = 1$.

The fact that we have but one equation doesn't mean we can't solve the problem. It only means there is more than one solution (in this case $x = cos(t), y = sin(t)$ is an infinite amount of solutions).

Answer 3

If you consider the problem "the number of equations >= number of unknown variables" in linear system of equations. The answer is The Kronecker-Capelli for linear system of equations.

Answer 4

Let $f:\mathbb{C}^m\rightarrow \mathbb{C}^n$ be a polynomial function. Thus $f(X)=0$ is a system of $n$ algebraic equations in $m$ unknowns. If $n<m$, then $\{X;f(X)=0\}$ is void or has an infinity of elements. The proof is very easy when we consider affine equations. It is much more difficult in the general case, cf. my post in:

How to prove rigorously that you need $m \geq n$ equations with $n$ unknowns to be able to solve a system of equations?