another proof of rank-nullity theorem

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

Proposition 1.

$\dim (V)=\mathrm{rank}(\varphi )+\mathrm{nullity}(\varphi )$.

Proof.

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

$$\dim (V)=\dim (K)+\dim (L)$$

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

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

$$\dim (V)=\dim (K)+\dim (L)=\mathrm{nullity}(\varphi )+\mathrm{rank}(\varphi ).$$

∎

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).