Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
What have you tried? –  Jonas Meyer Jun 6 '12 at 10:55
    
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

2 Answers 2

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

share|improve this answer
    
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 $\phi(V)=\phi^2(V)$ ]
share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.