Relation between the elementary Rouché-Fröbenius theorem and the more abstract Fröbenius theorem with exact sequences.

Question

In highschool I was taught that the "Rouché-Fröbenius theorem" was that given a homogenous linear system of equations, the solution space has dimension $n-r$ where $n$ is the dimension of total space (number of variables) and $r$ is the number of linearly independent equations. (And another part about existence of a solution space if the system is non-homogenous, but that's not the point now.)

Then in the first course in Linear Algebra in college, I was taught the Fröbenius theorem (named only after Fröbenius this time? I assume the Rouché part is about a non-homogenous linear system, when the solution is an affine space) as follows:

If

$0\rightarrow E_1\xrightarrow i E_2\xrightarrow \pi E_3\rightarrow 0$

is an exact sequence of vector spaces, then

$0\rightarrow (E_3)^*\xrightarrow \pi^* (E_2)^*\xrightarrow i^* (E_1)^*\rightarrow 0 $,

with the corresponding transpose (dual) functions, is also exact. Today I was told they are the same theorem, stated differently. I'm having trouble seeing the equivalence, I think it has to do with the anihilator of a subspace being the dual of the quotient of the ambient space modulo that subspace.

Please forgive my formatting and my using of words instead of writing in latex. I don't know Latex properly.

Answer 1

Let $k$ be our field, $(\phi_i)_{i=1}^m$ functionals on $k^n$ representing our equations (specifically, we define $\phi$ so that: $a_1x_1+a_2x_2+\cdots+a_nx_n=\phi(x_1,\cdots,x_n)$, i.e. $\phi\sim[a_1\,\,a_2\,\,\cdots\,\,a_n]$ as a row vector in the canonical basis).

We have an exact: $$0\to V\to k^n\overset{(\phi_i)_{i=1}^m}{\longrightarrow}k^m$$And can define $U$ as the image of the last map. By elementary linear algebra, $\dim_kU=r$; $K$ here is defined as the kernel which is exactly the solution space to the system of equations. We want to calculate $\dim_k V$.

We are told that the short exact $0\to V\to k^n\to U\to0$ dualises to a short exact sequence $0\to U^\ast\to(k^n)^\ast\to V^\ast\to0$. Since the dimensions of the first two objects are $r$ and $n$, their quotient $V^\ast$ has dimension $n-r$; this requires a proof which is so similar to the usual proof of rank-nullity (which is basically what you call Rouche-Frobenius) that I think this whole exercise is pointless, but anyway, we find $\dim_kV=\dim_kV^\ast=n-r$ as desired.

We cannot so immediately go in the converse, since you haven't stated Rouche-Frobenius for infinite dimensional spaces/systems. But if your vector spaces are all finite-dimensional, then some $0\to E_1\to E_2\to E_3\to0$ which is short exact (with dimensions $r,n,n-r$ respectively) dualises to some $0\to E_3^\ast\to E_2^\ast\to E_1^\ast$ which is exact, and we must just check the last map surjects. Alternatively, we check the subspace $E_2^\ast/E_3^\ast\subset E_1^\ast$ is full; counting dimensions, we find $n-(n-r)=r$ as the dimension of $E_2^\ast/E_3^\ast$ and $r$ as the dimension of $E_1^\ast$ (using your Rouche-Frobenius, but also this lemma about the dimension of a quotient space) hence $E_2^\ast/E_3^\ast=E_1^\ast$ is forced; $E_2^\ast\to E_1^\ast$ indeed surjects.