Consider a parameter space Rn, comprising points a=[a1, a2,...an]'∈Rn. Suppose we've got two constrained sets within the space.
The first is linearly constrained: f(a1, a2,...an)=0, with f() being a linear function. So it's a linear subspace.
The second is nonlinearly constrained: g(a1, a2,...an)=0, with g() being a nonlinear function including only polynomial operations i.e. addition and multiplication. So it's a nonlinear manifold.
If the dimensions of the two sets are identical, then the intersection must be a null set of Lebesgue measure 0, comprising only countable crossovers with lesser dimension.
Am I right? How to prove it?
I afraid, there should be some additional restrictions on the polynomial. For example, the following non-linearly constrained set
(a1)2 - (a2)2 = 0
consists of two linear subspaces
a1+a2 = 0 and a1 - a2 = 0,
so it intersects with the linear subspace a1+a2 = 0 by the whole subspace. So, you must exclude such a possibility.
The branch of mathematics that works with intersections of zero sets of polynomials (which are called "algebraic manifolds") is algebraic geometry. You should ask advise from people that work in that area or from algebraists what the condition on the polynomial can ensure that its null set does not to contain a linear subspace of the same dimension.
I afraid, there should be some additional restrictions on the polynomial. For example, the following non-linearly constrained set
(a1)2 - (a2)2 = 0
consists of two linear subspaces
a1+a2 = 0 and a1 - a2 = 0,
so it intersects with the linear subspace a1+a2 = 0 by the whole subspace. So, you must exclude such a possibility.
The branch of mathematics that works with intersections of zero sets of polynomials (which are called "algebraic manifolds") is algebraic geometry. You should ask advise from people that work in that area or from algebraists what the condition on the polynomial can ensure that its null set does not to contain a linear subspace of the same dimension.
I dont think its possible. Linear and non linear aspects cannot go together.
Dear Ning:
You are posing the wrong dimension condition. It does not matter whether or not the dimensions of the two manifolds $M_1, M_2$ are equal. What matters is the dimension of $M_1$ and the codimension of $M_2$, like for two linear subspaces $V_1,V_2$: Their intersection is a linear subspace of dimension at least $dim V_1 - codim V_2$ (according to the formula from linear algebra, $dim (V_1+V_2) = dim V_1 + dim V_2 - dim (V_1 \cap V_2)$). The same formula holds for algebraic manifolds once we complexify and projectivize (if we don't, the intersection is even smaller), where linear subspaces are replaced by algebraic varieties (Chow's theorem in algebraic geometry, see https://en.wikipedia.org/wiki/Chow's_theorem). In your case, however, the situation is much easier since $M_1$ is linear. Therefore you can consider the other (non-linear) equation just as an equation in $M_1$ (using the linear equations for $M_1$ to remove the additional variables).
Best regards
Jost
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