Let me first say I've been self-studying the implicit function theorem in the last few weeks, but also that my knowledge in linear algebra is still poor. Until now, I've seen the generalizations of the theorem, several examples, 'til the Jacobian.
Now, I've come across this interesting pdf, in which it is stated: "If a function $F$ is vector-valued, a...
If a function $F$ is vector-valued, a regular point is one where the total derivative (matrix) has linearly independent rows.
Why? Probably I'm not visualizing the concrete situation, but I cannot figure out in my mind what's going on... Ok, let's say a vector-valued function defines a system of two scalar functions whose outputs together form a vector itself (?). The gradients of these two functions are linearly independent. But what does that "physically" mean, and especially how does that relate to the implicit function theorem? I mean, why if those gradients are independent and the other conditions of the theorem are satisfied, then it is possible to write a variable as a function of the others?
I derivated this result considering a system of two three-variable functions $F$ and $G$ and found that it is possible to write locally $(x,y,z)$ as $(x,α(x),β(x))$ if the determinant of the 2 by 2 matrix of the partial derivatives of $F$ and $G$ with respect to $y$ and $z$ is non zero. Which means that the two gradients are linearly independent.
In fact, once we obtain
we ask that this
must be different from zero, but this is in fact a condition on the determinant:
But how? What is the extraordinary link between the theroetical and calculus-based conclusion of before and the determinant of the Jacobian? I feel I'm missing this point.