Homomorphisms and the Rank-Nullity Theorem
Most people who have taken an introductory Linear Algebra Course will be familiar with the Rank-Nullity Theorem. This article presents a strikingly concise proof to this theorem which is rarely seen in such courses.
Theorem
Let $V$ and $W$ be vector spaces and $T:V\rightarrow W$ be a linear map. Then, the following statement is true:
$$\dim{V} = \dim{\mathrm{Im}{T}} + \dim{\mathrm{Ker}{T}}$$
Proof
We let a set $V$ be a vector space. We note that $(V, \oplus)$ forms an abelian group. Furthermore, we note that the linear map $T$ is a homomorphism. Therefore, by the first fundamental theorem of homomorphisms:
$$
\begin{aligned}
V/\mathrm{Ker}{T} &\cong \mathrm{Im}{T}\\
\dim(V/\mathrm{Ker}{T}) &= \dim{\mathrm{Im}{T}} \\
\dim{V} - \dim{\mathrm{Ker}{T}} &= \dim{\mathrm{Im}{T}}\\
\end{aligned}
$$
Our result follows immediately.