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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.

Sam
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    First question: $\ker(T)\neq 0$ implies that there exists a nonzero $v\in \ker(T)$. You have used the wrong negation $v\not\in \ker(T)$. – Dietrich Burde Sep 28 '19 at 14:44

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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:

  • $V$ is the direct sum of $\mathrm{span}(v)$ and a corresponding complementary space $V'$: $$V = V' \dot + \mathrm{span}(v)$$
  • $v \color{blue}{\in} \mathrm{ker}(T)$ and $v \neq 0$

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 $$n-1=\mathrm{dim}(V')=\mathrm{dim}(\mathrm{ker}(\color{blue}{T|_{V'}}))+\mathrm{dim}(\mathrm{im}(T_{V'}))$$ $$ \stackrel{v \in \mathrm{ker}(T),v \neq 0}{=} \mathrm{dim}(\mathrm{ker}(\color{blue}{T}))\color{blue}{-1}+\mathrm{dim}(\mathrm{im}(\color{blue}{T}))$$

trancelocation
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  • I see. Although now I'm confused about $n-1=\mathrm{dim}\mathrm{ker}(T_{V'})+\mathrm{dim}\mathrm{im}(T_{V'})=\mathrm{dim}\mathrm{ker}(T)+\mathrm{dim}\mathrm{im}(T)=n$ – Sam Sep 28 '19 at 15:59
  • Just a few moments .. I will add something to my answer addressing this. – trancelocation Sep 28 '19 at 16:10
  • @Leo Just added some additional lines. Basically, one reduces the dimension of the kernel of $T$ by $1$ which is possible since the kernel is non-trivial. Then, one considers the same mapping but restricted to the subspace $V'$ (which now has a kernel with a dimension reduced by $1$). – trancelocation Sep 28 '19 at 16:33