Lagrange paper bug

Lagrange paper bug PhL 9.20.16 This paper has been out for quite a while now. I did not take time to really check out its logic too much, and today I have a question. In my proof of Theorem 1 for my sample case, I start by "eliminating" the three variables x1, x2, x3. I then reach this point where now I can express all four of these equations in terms of just the residual variables. I then go on to look at derivatives and I end up with three 4x4 matrix equations. I then claim that all three of the matrices must have zero determinant otherwise we get a contradiction. In this way, I go part way on my proof that all 4x4 submatrices of the R matrix must have zero determinant. When I repeat this eliminating all other possible sets of three variables, I claim to have shown that all 4x4 submatrices are zero and therefore the R matrix has rank < 4 and this is then associated with df = 0 since df = 0 was used to obtain some of the equations. But how do you know you can do this elimination? For example, how do you know you can eliminate x1, x2, x3 for some arbitrary set of 3 constraint equations? I suggest this can always be done, but I suspect it cannot always be done. Let's just look first at some simple cases. 1. Suppose one of the three constraints is x1 = 1. Then you have X1(x4, x5, x6) = 1, so that is no problem. 2. Suppose a constraint is x1 + x2 = 0. You could eliminate x1 everywhere it appears elsewhere by setting it equal to -x2. So then I guess you have x1 = - X2(x4, x5, x6) and that would be OK. What do I know in general about a set of N weird equations in M unknowns for M > N? Have I ever addressed this pure math question? Go back to the general case of a(x1,x2....xN) = 0. This equation describes an N-1 dimensional surface in EN, as I note on page 4. How do I know this fact? I know that a1dx1+a2dx2 + ... + aNdxN = 0. Well near (4.1.3) I claim that this surface fact is true, and moreover I claim that a is normal to this surface at any point on the surface. But how do I know it is a surface? If I knew that I could solve the equation for just one of the variables xs, then I could say xs = Xs(x1....xn) where xs is missing from the list. This surely is the equation of a surface in En. Of course if x2 did not appear in a(x1,x2....xN) = 0, I certainly could not solve the equation for x2 and make my surface statement based on X2. But I know that at least some xs appears in the equation. Maybe better to look at da = a1dx1+a2dx2 + ... + aNdxN = 0. There must be some derivative which is non-zero in this equation. Let xs be such a term. Then solve to get dxs = [a1dx1+a2dx2 + ... + andxn]/as // asdxs is missing from the sum This equation then says that if you move very slightly in all those other variables, a dxs is implied, and if you move with this dxs then you will stay on the surface which is defined by da = 0. So as long as none of the derivatives is infinite, this means you have a piece of surface. Question: Consider then a smooth surface a(x1,x2....xn) = 0. Can you solve this for x1 ? There are several cases: (a) Suppose x1 does not appear in the equation. That means that the surface is an extruded surface along this axis. It is the extrusion of the surface a(x2.....xn) of one lower dimension. You solution for x1 is then that any value of x1 is a solution. So you could take x1 = 1 as your solution. (b) otherwise, I would argue that locally at some point the surface is linear and as such, the surface is a plane in En defined by a1dx1+a2dx2 + ... + aNdxN = 0. This is the tangent plane. OK, now consider these two equations: a(x1,x2....xN) = 0 b(x1,x2....xN) = 0 each of which describes a surface in En. The intersection of these surfaces is a surface of dimension n-2 in the space En. Consider some point r which lies on this dim n-2 surface. Near that point both those surfaces are linear, so this is the intersection of two dim n-1 planes. Question: If you intersect two dim n-1 planes in En , you get an n-2 dim plane. If we select a point on the intersection as our coordinate origin, we have (vectors are derivative vectors, and r means dr ) a r = 0 b r = 0 a1x1+a2x2 + ... + aNxN = 0. b1x1+b2x2 + ... + bNxN = 0. Can you solve these equations for x1 and x2 ? Here are the cases: (a) x1 and x2 appear in neither equation. Then solution is x1 = 0 and x2 = 0, fine. (b) x1 is missing from one of the equations. Solve that equation for x2 = X2(x3, ....xn). Then put this into the other equation for x2 and solve that equation for x1. Done. I have dealt with this little problem somewhere else recently! Where was it? The issue is the solvability of M linear equations in N unknowns and involves Cramer's rule. I search docs for Cramer and unknowns and I get a hit in my non-square matrix algebra stuff. I think this is the idea. In the above equations, assume that x3 ....xN have some assigned values. Then our two equations are a1x1+a2x2 = - stuff1 b1x1+b2x2 = - stuff2 Regardless of the values of stuff1 and stuff2, you can Cramer-solve these equations for x1 and x2 as long as det ≠ 0 That is, the rows have to be linearly independent. We are talking here about the intersection of two lines in E2. The only way there could NOT be a solution for x1,x2 is if the lines are parallel which means that the normals are parallel which means that a = k b and so the rows are linearly dependent. I am feeling better now that you really can do what I claim in Lagrange doc.