Let $E$ be a $\Bbb{C}$ vector space of dimension $n$ and let $u$ be a linear application ($\mathcal{L}(E)$ on $E$. How can one prove that $u$ is a homotethy iff $\forall v,w\in\mathcal{L}(E),u=v\circ w\Rightarrow u=w\circ v$ ?
My thoughts :
for the forward case : there exists $\lambda\in\Bbb{C},e_1\in E$ such that $u(e_1)=\lambda e_1$ ($u(E)=span(e_1)$). If we create a basis $(e_1,\dots,e_n)$ of $E$, $u(e_{j>1})=0$. Let $v,w$ such that $u=v\circ w$. Thus $v\circ w(e_i)=\lambda e_i,w(e_{j>1})\in\ker v$ ....
for the backwards case : ? Maybe we could suppose that there exists $v,w$, one of which is non invertible, such that $u=v\circ w$. Since $u(e_{j>1})=0$, we could reason on their $\ker$ and $Im$, but I haven't been able to link that to $u$ being a homotethy ...
If $0_n$ is a homothety, then the implication $0=v\circ w\Rightarrow 0=w\circ v$ is false. We define a homothety as $u\in L(E)$ s.t. there is $\lambda\in\mathbb{C}^*$ with $u=\lambda I_n$. Then $v\circ w=\lambda I_n$ implies that $v=\lambda w^{-1}$ and $v,w$ commute.
Conversely, assume that $u$ is not a homothety. Or $u=0$ and the above implication is false either there are non-parallel vectors $x,y$ s.t. $u(x)=y$. Let $F$ s.t. $E=span(x,y)\bigoplus F$. We define $v(z)=u(z),w(z)=z$ when $z\in F$ and $v(y)=y,v(x)=u(y),w(x)=y,w(y)=x$. Then $v\circ w=u$ and $w\circ v\not= u$ (except perhaps if $u(y)=x$; in this case, it remains to find a variant for the choice of $v,w$).