Let $V$ be a finite dimensional $\mathbb{K}$-vector space and $\varphi: V \to V$ a linear mapping. Show that the following statements are equivalent.
\begin{align}
i)&V=\operatorname{ker}\varphi\oplus \operatorname{im} \varphi\\
ii)&\operatorname{ker}\varphi=\operatorname{ker}(\varphi\circ \varphi)\\
iii)& \operatorname{im}\varphi=\operatorname(\varphi \circ \varphi)
\end{align}
$i)$ Definition of direct sum says, that $\operatorname{ker}\varphi\cap\operatorname{im} \varphi=\{0\}$. Therfore, $0_V\in\operatorname{ker}\varphi$ and $0_V\in\operatorname{im}\varphi$ This implies that: $$\operatorname{ker}\varphi=\operatorname{ker}(\varphi\circ \varphi)=\operatorname{ker}(\varphi)\circ \operatorname{ker}(\varphi)=0\circ 0=0=\operatorname{ker}\varphi$$ and $$\operatorname{im}\varphi=\operatorname{im}(\varphi\circ \varphi)=\operatorname{im}(\varphi)\circ \operatorname{im}(\varphi)=0\circ 0 = 0=\operatorname{im}\varphi$$
All statements are equivalent.
Is this proof correct or do I need to show $i\to ii, ii\to i, ii\to iii, iii\to ii, i\to iii, iii\to i$ etc.?
Best Answer
For $i)\implies ii)$
We have trivially $\ker\varphi\subset \ker (\varphi\circ\varphi)$. Conversely, let $x\in \ker (\varphi\circ\varphi)$ so $\varphi(x)\in \ker\varphi\cap {\rm im}\varphi=\{0\}$ so $x\in\ker\varphi$.
For $ii)\implies iii)$
Use the fact ${\rm im}(\varphi\circ\varphi)\subset {\rm im}(\varphi)$ and the rank-nullity theorem.
For $iii)\implies i)$
Let $x\in\ker\varphi\cap{\rm im}\varphi$ then for somme $x'$ we have $x=\varphi(x')$ and so $x'\in\ker(\varphi^2)$. Again we have from rank-nullity theoren that $x'\in\ker\varphi$ and so $x=0$. Conclude the desired result using again the rank-nullity theorem.