In order to find $n$ unknowns why do we need $n$ equations (non linear case)

Question

As indicated by the title, why do we need at least $n$ equations (linearly independent?) to find $n$ unknowns?

Does this rule differ between linear case and non-linear case?

I found this related question: Necessary condition for uniqueness solution in a system of non-linear equations , however I am asking about the existance of a solution and not the uniqueness.

Edit : To make my quetsion more specific: I am having $n$ unknowns $x_1, \cdots, x_n $ and a generating equation $ f(x_1, \cdots, x_n,t)= 0$. $f$ here is a polynomial in $x_1, \cdots, x_n,t$. As the function $f$ depends on $t$, so if I change $t$ I get a new equation involving $x_1, \cdots, x_n $. I want to find the values of these $n$ knowns. So how many equations I need ( generated from varying $t$ in $f$) to gauranre a solution for $x_1, \cdots, x_n $ .

Giving some references in this manner would be highly appreciated.

Thank you in advance

Answer 1

Think of it as the intersection of hyper-surfaces. To begin with keep it simple and stick to the linear case in 2-d. So two unknowns and two equations. Each equation defines a line in a 2-d space. At some point they will intersect and this point is the solution. (If the lines are parallel then the equations are not independent and hence no solution.) Now if the equations were non-linear then they may not intersect, and hence no solution can be found.

Going back to the linear case but now in three dimensions. An equation now defines a plane in 3-d space. Two planes will intersect to give a line. The third plane will intersect this line to give a point which is the solution. In higher dimensions its just more of the same.

Answer 2

It's difficult to prove rigoruosly, but here is an intuitive idea, based on the notion that a single equation with more than one variable has too many solutions for our taste, a single equation in one unknown has a good number of solutions, while several equations in a single variable has too few solutions (usually none).

Say we have our equations $$ \cases{f_1(x_1, x_2, x_3,\ldots,x_n) = 0\\ f_2(x_1, x_2, x_3,\ldots,x_n) = 0\\ f_3(x_1, x_2, x_3,\ldots,x_n) = 0\\ \vdots\\ f_m(x_1, x_2, x_3,\ldots,x_n) = 0 } $$ Now, we solve the first equation with respect to $x_1$ to get something of the form $x_1 = g_1(x_2, x_3,\ldots,x_n)$ (in other words, we get $x_1$ alone on one side, and no $x_1$'s on the other side). This we can now insert into all the other equations. We have now used the first equation, so it is useless as far as solving the system goes (it tells us what $x_1$ is once we've solved all the others, but there are no $x_1$ in the other equations left, so it doesn't tell us anything that helps with the other equations).

Thus we've used up one equation to eliminate one variable. We can keep going, using one equation to eliminate one variable until we get to the last one. Now, if $m<n$, then we are on the last equation, we've used all the others, but we still have more than one variable left. Thus the system has too many solutions for our taste. If, on the other hand, $n<m$, then we are left with several equations, all in one variable, which is not good either. We are left with the case $m = n$, in which case we are left with one equation with one variable, which is exactly what we wanted.

Note about non-linearity: If, say, the first equation is non-linear, then there might be several solutions for $x_1$. In other words, there might be several $g_1$'s that solve the equation. In that case you split the system of equations into one for each $g_1$. For instance, if the first equation is quadratic, so you get two solutions, then you end up with two different sets of $m-1$ equations in $n-1$ unknowns. They should be solved separately.

Also, introducing higher degrees makes it possible that you have no solutions, even though the equations are independent and there are as many equations as there are unknowns. This is just like a regular one-variable equation in higher degree. For instance $$ \cases{x^2 + y^2 = 1\\ x^2 - y = 2} $$ has no solutions. On the other hand, if you change that last equation's right-hand side to $= 1.1$, you get four solutions, if you change it to $= 1$ you get three solutions, if you change it to $= 0$ you get two solutions, and if you change it to $= -1$ you get one solution.