For the Lagrange Multiplier Theorem, why must the number of constraints be less than the number of variables?

Question

In Fitzpatrick's "Advanced Calculus", the beginning of the General Lagrange Multiplier Theorem is given by:

Let $U$ be an open subset of $\mathbb{R}^n$ and suppose that $f: U \rightarrow \mathbb{R}$ is continuously differentiable. Let $k$ be a positive integer less than $n$ and suppose that the mapping $G: U \rightarrow \mathbb{R}^k$ is also continuously differentiable."

The function $G$ is a vector valued function, where each component function is a constraint. Why must $k < n$, that is, why must the number of constraints be less than the number of variables?

Answer 1

I guess here the constraints are equalities given by $G(x)=0$. As a concrete example to @amsmath's answer, just think about the case where $G$ is a linear mapping, then the constraints are $k$ linear equations in $n$ variables. If $k>n$, then some constraints are linearly dependent, therefore, are redundant and can be ignored (of course when the problem is feasible).

The method of Lagrangian multiplier also applies to the inequality case, where the constraints involve inequalities such as $G(x)\le 0$, then $k$ definitely does not have to be less than $n$. See, for example, the KKT condition in nonlinear optimization https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions

Answer 2

Each constraint defines a hypersurface, which is by definition of dimension $n-1$. Hence, $k$ constraints define an intersection of $k$ such hypersurfaces, which is a manifold of dimension $n-k$.