I recently stumbled across this Proof of the Rank Nullity Theorem, and there is a step in the Induction Hypothesis part of the proof which I do not understand.
Induction Hypothesis: Assume the theorem holds for $\mathrm{dim}(V)=n-1$.
Let $T:V\rightarrow W$ be a linear transformation with $\mathrm{dim}(V)=n$. If $\mathrm{ker}(T)=\{0\}$ then $T$ is a bijection $V\rightarrow \mathrm{im}(T)$ and the claim is true. If $\mathrm{ker}(T)\ne \{0\}$ then there exists $v\notin \mathrm{ker}(T)$ with $v\ne 0$. Let $V'=V-\mathrm{span}(v)$
$$n-1=\mathrm{dim}(V')=\mathrm{dim}(\mathrm{ker}(T|_{V'}))+\mathrm{dim}(\mathrm{im}(T_{V'}))=\mathrm{dim}(\mathrm{ker}(T))+\mathrm{dim}(\mathrm{im}(T_{V'}))$$
by induction, thus
$$n=\mathrm{dim}(V)=\mathrm{dim}(V'+\mathrm{span}(v))=\mathrm{dim}(\mathrm{ker}(T))+\mathrm{dim}(\mathrm{im}(T)))$$
How does he get to the last line? In particular, why does the $\mathrm{dim}(\mathrm{im}(T|_{V'}))$ in the second to last line become $\mathrm{dim}(\mathrm{im}(T))$?
Would appreciate any help.
Maybe your confusion comes from writing "$V'=V-\mathrm{span}(v)$" and from the erroneous "$v\notin \mathrm{ker}(T)$".
For the understanding of the induction step the following two facts are important:
Then the missing part in your proof is $\mathrm{im}(T|_{V'}) = \mathrm{im}(T)$. This is easily seen as follows using the fact that each $x \in V$ can be written as $x=v'\dot +\lambda v$ with $v' \in V'$ $$y =Tx \stackrel{x=v'\dot +\lambda v}{\Longrightarrow}y= Tv' + \lambda\underbrace{Tv}_{=0} = T|_{V'}v'$$ $$\Rightarrow y \in \mathrm{im}(T|_{V'})$$
Edit addressing open question in comment:
We have $$n-1=\mathrm{dim}(V')=\mathrm{dim}(\mathrm{ker}(\color{blue}{T|_{V'}}))+\mathrm{dim}(\mathrm{im}(T_{V'}))$$ $$ \stackrel{v \in \mathrm{ker}(T),v \neq 0}{=} \mathrm{dim}(\mathrm{ker}(\color{blue}{T}))\color{blue}{-1}+\mathrm{dim}(\mathrm{im}(\color{blue}{T}))$$