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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 $\phi(V ) = \phi^2(V)$

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To show that $V = A \oplus B$ for subspaces $A$, $B$, you need to show (1) that every vector $v \in V$ can be written as $v = a + b$ for some $a \in A, b \in B$ and (2) that $A \cap B = \{ 0 \}$. – Adeel Jun 6 '12 at 12:57
up vote 1 down vote accepted

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

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if I now say let $a\in\phi^{2}(V)$ therefore $b\in V$ s.t $b=\phi(c)+d$ for $c \in V$ and $d \in NS(\phi)$ therefore $a=\phi(b)$ is this correct – sarah jamal Jun 19 '12 at 20:15

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 $\phi(V)=\phi^2(V)$ ]
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I think you meant it is clear that $\phi(V) + NS(\phi) \subseteq V$. – André Caldas Jun 6 '12 at 15:26
Yes that is what I mean, I'll edit to fix. – Nate Iverson Jun 6 '12 at 19:13

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