another proof of rank-nullity theorem

Proposition 1.

$\operatorname{dim}(V)=\operatorname{rank}(\phi)+\operatorname{nullity}(\phi)$.

Proof.

Let $K=\operatorname{ker}(\phi)$. $K$ is a subspace   of $V$ so it has a unique algebraic complement $L$ such that $V=K\oplus L$. It is evident that

 $\operatorname{dim}(V)=\operatorname{dim}(K)+\operatorname{dim}(L)$

since $K$ and $L$ have disjoint bases and the union of their bases is a basis for $V$.

Define $\phi^{\prime}:L\to\phi(V)$ by restriction  of $\phi$ to the subspace $L$. $\phi^{\prime}$ is obviously a linear transformation. If $\phi^{\prime}(v)=0$, then $\phi(v)=\phi^{\prime}(v)=0$ so that $v\in K$. Since $v\in L$ as well, we have $v\in K\cap L=\{0\}$, or $v=0$. This means that $\phi^{\prime}$ is one-to-one. Next, pick any $w\in\phi(V)$. So there is some $v\in V$ with $\phi(v)=w$. Write $v=x+y$ with $x\in K$ and $y\in L$. So $\phi^{\prime}(y)=\phi(y)=0+\phi(y)=\phi(x)+\phi(y)=\phi(v)=w$, and therefore $\phi^{\prime}$ is onto. This means that $L$ is isomorphic   to $\phi(V)$, which is equivalent      to saying that $\operatorname{dim}(L)=\operatorname{dim}(\phi(V))=\operatorname{rank}(\phi)$. Finally, we have

 $\operatorname{dim}(V)=\operatorname{dim}(K)+\operatorname{dim}(L)=% \operatorname{nullity}(\phi)+\operatorname{rank}(\phi).$

Remark. The dimension of $V$ is not assumed to be finite in this proof. For another approach (where finite dimensionality of $V$ is assumed), please see this entry (http://planetmath.org/ProofOfRankNullityTheorem).

Title another proof of rank-nullity theorem AnotherProofOfRanknullityTheorem 2013-03-22 18:06:14 2013-03-22 18:06:14 CWoo (3771) CWoo (3771) 4 CWoo (3771) Proof msc 15A03 ProofOfRankNullityTheorem