Working with Linear Operators on Vector Space

Question

Let $\phi: V \rightarrow V$ be a linear operator on a vector space V over a field F

Prove that $V = \phi(V)\bigoplus NS (\phi )$

if and only if $\phi(V ) = \phi^2(V)$

Answer 1

I'll use $\,\ker\phi\,$ instead of $NS$:

Suppose $\,V=\phi(V)\oplus\ker\phi\,$ and let$\,x\in\phi(V)\Longrightarrow\,\exists y\in V\,\,s.t.\,\,x=\phi y$ , but:$$y\in V\Longrightarrow\,\exists!\,v=\phi t\in\phi(V)\,,\,u\in\ker\phi\,\,s.t.\,\,y=\phi t+u\Longrightarrow$$$$\Longrightarrow x=\phi y=\phi^2t+\phi u=\phi^2t\Longrightarrow x\in\phi^2(V)$$Now you try the other direction

Answer 2

Here's a sketch of what you need to do:

1. Show if $V = \phi(V)\bigoplus NS (\phi )$ then $\phi(V)=\phi^2(V)$.
   This is easy think about applying $\phi$ to a generic element of $\phi(V)\bigoplus NS (\phi )$

2. Show if $\phi(V)=\phi^2(V)$ then $V = \phi(V)\bigoplus NS (\phi )$

   • It should be clear that $\phi(V)+ NS (\phi ) \subseteq V$.
   • Show $V=\phi(V)+NS(\phi)$.
   • Show $\phi(V)\cap NS(\phi)=\{0\}$ [This comes from $\phi(V)=\phi^2(V)$ ]