# The dimension theorem and pedagogy

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.

• I think that this is the kind of question which is most appropriate for the Math Educators SE. I am going to migrate it there. – Xander Henderson Oct 22 '20 at 12:51
• @Xander Henderson Thank you so much. – Boka Peer Oct 22 '20 at 12:54
• Any reason for specifying an undergraduate junior? (Third year of a four-year Bachelor's degree in the US and some other countries,) – J W Oct 22 '20 at 14:02
• I have tried to mean that the students who don't have much mathematics background. Previously I have seen that some junior/seniors take their first proof-based linear algebra class. I am sorry for not clarifying things I intended to mean. Thank you. – Boka Peer Oct 23 '20 at 4:24

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).