Studying for my finals in calculus 3, returning to the proof of Lagrange multipliers with multiple constraints, I'm having a hard time getting any form of intuition about why this is the case - why does this give us the extrema points.
While looking at one constraint, I understand the geometric interpretation of the gradient vectors being linearly dependent. And I'm able to understand that each constant defines a surface one dimension lower since we are looking at the group defined by intersecting all: $\forall i\in [n]: g_i(x)=0$
If my function $f$ is linearly dependent on all other constraining functions, what does this mean visually(what does it mean visually that all other constraints are independent and $f$ is dependent on all of them)?
Why is it necessary for all gradients of $g_i$ to be linearly independent? (or is this just the version of the proof I was shown, and a stronger argument exists? I'll add that we proved the Theorem using the Open Mapping Theorem).
Additions
I can see why (2) is required since we want to think of the simplified version with one constraint, which is only possible when the intersection defines some curve. I still am unsure if this is required since we know we can find local extrema on open groups and are not limited to a single curve. Or is the whole concept trying to avoid this method and use the dependency of gradients to optimize?
A counterexample for your question 2.
Consider the problem:
$\min x_{2}$
subject to
$(x_{1}-1)^{2}+x_{2}^{2} \leq 1$
$(x_{1}+1)^{2}+x_{2}^{2} \leq 1$
It's easy to see that the only feasible solution is $x_{1}=x_{2}=0$. but there are no Lagrange multipliers that work for that point. Here the constraint gradients are linearly dependent, so the theorem isn't violated.
There are many alternative hypotheses, called "constraint qualifications" that can be used instead of the Linear Independence Constraint Qualification (LICQ). Whole books have been written about alternative hypotheses.