Proof of The Rank–Nullity Theorem

Routine facts about mutual inclusion of classes are used without comment (Toolkit for AL2). The kernel $\ker T$ is a subspace of $V$ (Kernel and Image of a Linear Map Are Subspaces (ii)), hence a vector space with the operations of $V$ (Linear Subspaces Are Vector Spaces; Intersections (i)), and it is finite-dimensional (Properties of Dimension (v)). Choose a basis $(u_{1},\dots,u_{m})$ of $\ker T$, with $m=\dim\ker T$ (Finite-Dimensional Vector Spaces and Their Dimension (ii)).

By Linear Subspaces Are Vector Spaces; Intersections (iv), the family $(u_{1},\dots,u_{m})$ is linearly independent also as a family in $V$, so it extends to a basis $(u_{1},\dots,u_{m},v_{1},\dots,v_{k})$ of $V$ (Properties of Dimension (i)), and $\dim V=m+k$ (Finite-Dimensional Vector Spaces and Their Dimension (ii)). A linear combination of this basis splits as $\sum_{i=1}^{m}a_{i}u_{i}+\sum_{j=1}^{k}b_{j}v_{j}$ (Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (ii), Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (viii)). We show that $(Tv_{1},\dots,Tv_{k})$ is a basis of $\operatorname{im}T$; this gives the theorem.

The image $\operatorname{im}T$ is a subspace of $W$ (Kernel and Image of a Linear Map Are Subspaces (ii)), hence a vector space with the operations of $W$ (Linear Subspaces Are Vector Spaces; Intersections (i)), and it contains every $Tv_{j}$. Let $w\in\operatorname{im}T$, say $w=Tv$ with $v\in V$ (Kernel and Image of a Linear Map (ii)), and write $v=\sum_{i}a_{i}u_{i}+\sum_{j}b_{j}v_{j}$ (Coordinates with Respect to a Basis (i)). The vector $\sum_{i}a_{i}u_{i}$ lies in $\ker T$, a subspace containing every $u_{i}$ (The Span of a Finite Family is the Smallest Subspace Containing It (iii)), so by Linear Maps, Linear Operators and Isomorphisms of Vector Spaces (i) and Kernel and Image of a Linear Map Are Subspaces (i)

$$w=T\Bigl(\sum_{i}a_{i}u_{i}\Bigr)+T\Bigl(\sum_{j}b_{j}v_{j}\Bigr)=0+\sum_{j}b_{j}\,Tv_{j}.$$

So $w$ lies in the span of $(Tv_{1},\dots,Tv_{k})$ in $W$, which is its span in $\operatorname{im}T$ (Linear Subspaces Are Vector Spaces; Intersections (iv)). Hence $(Tv_{1},\dots,Tv_{k})$ spans $\operatorname{im}T$, and $\operatorname{im}T$ is finite-dimensional (Finite-Dimensional Vector Spaces and Their Dimension (i)).

It remains to show that $(Tv_{1},\dots,Tv_{k})$ is linearly independent. Let $\sum_{j}c_{j}Tv_{j}=0$. Then $T(\sum_{j}c_{j}v_{j})=0$ by Kernel and Image of a Linear Map Are Subspaces (i), so $\sum_{j}c_{j}v_{j}\in\ker T$. As $(u_{1},\dots,u_{m})$ spans $\ker T$, and its span in $\ker T$ is its span in $V$ (Linear Subspaces Are Vector Spaces; Intersections (iv)), there are $d_{1},\dots,d_{m}\in K$ with $\sum_{j}c_{j}v_{j}=\sum_{i}d_{i}u_{i}$. By The Span of a Finite Family is the Smallest Subspace Containing It (i), Elementary Identities in a Vector Space (iv) and Elementary Identities in a Vector Space (vii), the linear combination of the basis of $V$ above with coefficients $(-d_{1},\dots,-d_{m},c_{1},\dots,c_{k})$ is

$$\sum_{i}(-d_{i})u_{i}+\sum_{j}c_{j}v_{j}=\sum_{j}c_{j}v_{j}-\sum_{i}d_{i}u_{i}=0_{V},$$

so every $c_{j}=0$ by the independence of that basis (Linearly Independent and Linearly Dependent Finite Families (i)). By Linear Subspaces Are Vector Spaces; Intersections (iv), $(Tv_{1},\dots,Tv_{k})$ is then linearly independent in $\operatorname{im}T$, so it is a basis of $\operatorname{im}T$ (Finite Basis of a Vector Space (i)), $\dim\operatorname{im}T=k$ (Finite-Dimensional Vector Spaces and Their Dimension (ii)), and

$$\dim V=m+k=\dim\ker T+\dim\operatorname{im}T.$$

When $m=0$ or $k=0$ the corresponding family is empty, and the argument applies unchanged: the empty family is linearly independent (Elementary Properties of Linear Independence (vi)) and its span is $\{0\}$ (The Span of a Finite Family is the Smallest Subspace Containing It (iv)).