15A03-AnotherProofOfRanknullityTheorem.tex

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\pmtitle{another proof of rank-nullity theorem}
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\begin{document}
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 $\phi$.

\begin{prop} $\dim(V)=\rank(\phi)+\nullity(\phi)$. \end{prop}

\begin{proof}  Let $K=\ker(\phi)$.  $K$ is a subspace of $V$ so it has a unique algebraic complement $L$ such that $V=K\oplus L$.  It is evident that $$\dim(V)=\dim(K)+\dim(L)$$ since $K$ and $L$ have disjoint bases and the union of their bases is a basis for $V$.  

Define $\phi': L\to \phi(V)$ by restriction of $\phi$ to the subspace $L$.  $\phi'$ is obviously a linear transformation.  If $\phi'(v)=0$, then $\phi(v)=\phi'(v)=0$ so that $v\in K$.  Since $v\in L$ as well, we have $v\in K\cap L=\lbrace 0\rbrace$, or $v=0$.  This means that $\phi'$ 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 $\phi'(y)=\phi(y)=0+\phi(y) =\phi(x)+ \phi(y) =\phi(v) =w$, and therefore $\phi'$ is onto.  This means that $L$ is isomorphic to $\phi(V)$, which is equivalent to saying that $\dim(L)=\dim(\phi(V))=\rank(\phi)$.  Finally, we have $$\dim(V)=\dim(K)+\dim(L)=\nullity(\phi)+\rank(\phi).$$
\end{proof}

\textbf{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 \PMlinkname{this entry}{ProofOfRankNullityTheorem}.
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