= Solution
Assume $\phi^2=0$. Then
$$
\operatorname{im}\phi\subseteq\ker\phi.
$$
Set $W_1=\operatorname{im}\phi$, choose a <direct-sum complement> $U$ such that $\ker\phi=U\oplus W_1$, and choose a complement $W_2$ such that $V=\ker\phi\oplus W_2$.
The restriction $\phi|_{W_2}:W_2\to W_1$ is injective because $W_2\cap\ker\phi=0$, and it is surjective because every image $\phi(v)$ equals $\phi(w_2)$ after decomposing $v=k+w_2$. It is therefore a <linear isomorphism>[isomorphism], so $\dim W_2=\dim W_1$ and $\phi(W_2)=W_1$. Also $\phi(U)=\phi(W_1)=0$. Hence
$$
\boxed{V=U\oplus W_1\oplus W_2}
$$
has all the required properties.
Back to article page
