Rank-Nullity Theorem

Rank-Nullity Theorem: Let $T:V\to W$ be a linear map between finite-dimensional vector spaces over the same field. Define

• $\ker(T)=\{v\in V:T(v)=0\}$ (kernel / null space),
• $\operatorname{im}(T)=\{T(v):v\in V\}$ (image),
• $\operatorname{nullity}(T)=\dim(\ker(T))$,
• $\operatorname{rank}(T)=\dim(\operatorname{im}(T))$, where $\dim(U)$ denotes the number of vectors in any basis of $U$.

Then

This theorem is the basic dimension-counting tool for linear maps. For example, $T$ is injective if and only if $\ker(T)=\{0\}$, and it is surjective if and only if $\operatorname{rank}(T)=\dim(W)$ (in finite dimensions). Standard proofs ultimately rely on the existence of bases, guaranteed in general by basis existence .

Proof sketch: Choose a basis $(k_1,\dots,k_m)$ of $\ker(T)$ and extend it to a basis $(k_1,\dots,k_m,v_{m+1},\dots,v_n)$ of $V$. Then $(T(v_{m+1}),\dots,T(v_n))$ is a basis of $\operatorname{im}(T)$: it spans by construction, and linear independence follows because any dependence lifts to an element of $\ker(T)$. Counting basis elements gives $n=m+(n-m)$, i.e. $\dim(V)=\dim(\ker(T))+\dim(\operatorname{im}(T))$.