The dimension theorem and pedagogy

Question

The dimension theorem (the rank-nullity theorem) can be explained in many ways. I consider it as a consequence of the first isomorphism theorem/splitting lemma. When I teach undergrad matrix-theoretic linear algebra, I start with the equation $Ax=b,$ and I tell my students that the dimension theorem basically says that the number of total variables equals the sum of the number of free variables and the number of "non-free" variables. They find this statement very easy. If I teach a "formal/proof-based" undergrad mathematics class, I tell my students that the dimension theorem basically tells us how much "stuff" we need to put inside the nullspaces to extend it to the given vector space.

Today I found a very good analogy: In some sense, the dimension theorem is the linear algebraic analog of the Pigeonhole Principle. Note that for any finite set $A,$ the function $f: A \rightarrow A$ is injective iff surjective iff bijective. It's a consequence of the Pigeonhole Principle. The dimension theorem gives a similar kind of conclusion for a finite-dimensional vector space $V,$ and any linear map $T: V \rightarrow V.$

Now, could you help me by providing a couple of more analogies that can be explained to an undergrad junior? Thank you so much. Please stay safe.

Answer 1

It seems to me that a discussion of how to make any function $f: A \rightarrow B$ into a bijection might be in order. First, we can deal with onto by replacing $B$ with $f(A)$. So, let $g: A \rightarrow f(A)$ be the function given by $g(x)=f(x)$ for each $x \in A$. Next, we may need to make the domain smaller, for each non-empty fiber $$f^{-1} \{ b \} = \{ a \in A \ | \ f(a)=b \} $$ we should select one point. Let the collection of all such points be $S$. Define $h: S \rightarrow f(A)$ by $h(x)=f(x)$ for each $x \in S$. If $x,y \in S$ and $h(x)=h(y)$ then $f(x)=f(y)$ hence there exists $b \in B$ for which $x,y \in f^{-1}\{ b \}$. But, by construction of $S$, we have $x=y$. That is, $h$ is injective.

The innocent little step of selecting one point is rather difficult for arbitrary functions. In contrast, for linear transformations each fiber of the domain is an affine subspace. In particular, $T: V \rightarrow W$ has $$ T^{-1}\{ w \} = \{ x \in V \ | \ T(x)=w \} $$ If $x,y \in T^{-1}\{ w \}$ then $T(x)=w=T(y)$ hence $T(y-x)=0$ and so $y-x \in \text{Ker}(T)$. That is, $y = x+z$ where $z \in \text{Ker}(T)$. Indeed, $$ T^{-1}\{ w \} = x + \text{Ker}(T) $$ where $T(x)=w$. This is a very interesting equation because it means all the fibers of $T$ are the same size as $\text{Ker}(T)$. Ok, it's better than that, $\mathbb{R}$ and $\mathbb{R}^2$ have the same size, but they have different dimension. The dimension of the fibers of a linear map are all the same. Of course, the only way $T$ can be injective is for the size of these fibers to shrink to a single element, in turn that means the kernel must be zero dimensional.

The linear algebraic proof that a linear map $T: V \rightarrow W$ has $$\text{dim}(V) = \text{dim}( \text{Ker}(T))+ \text{dim}( \text{Image}(T))$$ is anchored to basis extension arguments. The basis for the kernel in $V$ is extended to those vectors outside the kernel. Then, those vectors outside the kernel have an image in $W$ which serves to generate $T(V)$. If $V=W$ then the only way the vectors outside the kernel can generate $T(V)=V$ is for there to be no vectors in the kernel (except zero).

I'm posting this in the hope that my answer might spur someone else to a better answer.