First Isomorphism Theorem of Vector Spaces

theoremvector-spaces

Intuition

This is exactly the analogue of Fundamental Isomorphism Theorem for groups, Fundamental Isomorphism Theorem (Ring) for rings, and every other first fundamental isomorphism theorem in algebraic theories.

Formal Statement

Let $T\in \mathcal{L}(V,W)$. Since $\mathrm{null}(T)$ is a subspace of $V$, we can define the quotient space ${}^V{/}_{\mathrm{null}(T)}$. Moreover,

$${}^V{/}_{\mathrm{null}(T)}\simeq \mathrm{range}(T)\subseteq W.$$

Proof

We want to describe an isomorphism $\tilde{T}:{}^V{/}_{\mathrm{null}(T)}\to \mathrm{range}(T)$.

Let us define $\tilde{T}(v+\mathrm{null}(T))=Tv$, or the coset of $v$ with respect to $\mathrm{null}(T)$ maps to an element $Tv$ in $\mathrm{range}(T)$.

(Injectivity) Assume $\tilde{T}(v+\mathrm{null}(T))=0$. Then by definition $Tv=0⟹v\in \mathrm{null}(T)⟹v-0\in \mathrm{null}(T)⟹v+\mathrm{null}(T)=0+\mathrm{null}(T)=$ zero of ${}^V{/}_{\mathrm{null}(T)}.$

(Surjectivity) Since $\tilde{T}(v+\mathrm{null}(T))=Tv$, it is obvious that $\mathrm{range}(\tilde{T})=\mathrm{range}(T)$, so $\tilde{T}$ is surjective.

Corollaries