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# another proof of rank-nullity theorem

Let $\phi:V\to W$ be a linear transformation from vector spaces $V$ to $W$. Recall that the rank of $\phi$ is the dimension of the image of $\phi$ and the nullity of $\phi$ is the dimension of the kernel of $\phi$.

###### 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.

## Mathematics Subject Classification

15A03*no label found*

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