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So let $V$ be a $\Bbb R-$ vector space of maybe unlimited dimension, and $f\in \textsf{End}(V)$.

I have to show that:

$f\circ f=f$ $\Longrightarrow$ $V=V_1\oplus V_0$

and:

$f \circ f=id$ $\Longrightarrow$ $V=V_1\oplus V_{-1}$

I just don't even really know how to start. I would appreciate any help, maybe the first one would already be enough and then I can try the second one alone.

2 Answers 2

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Hints: For the first, note that any vector may be decomposed as $$ v = f(v) + (v - f(v)) $$ For the second, we have $$ v = \frac 12 (v + f(v)) + \frac 12 (v - f(v)) $$ or alternatively, note that $g = \frac {f + id}{2}$ satisfies $g \circ g = g$.

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    I can't really understand the solution for second one.. could you maybe please explain it a bit more? – 2017-02-02
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    @user406473 do you see that $v + f(v) \in V_1$? That $v - f(v) \in V_{-1}$? – 2017-02-03
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    I think I got it now thank you – 2017-02-03
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Let me give you some hints. Think of it this way:

$ f^{2} = f$

means that $f$ acts trivially on its image $W$. If $K = ker (f)$, then $W$ and $K$ are disjoint. Now if $w \notin ker(f)$, then $w$ is surely in the image? Can you find its preimage?

For the second problem, clearly, $f(v) = v, -v$ are two options that come to mind. You have to prove that these are the only two. Clearly they are disjoint and now prove that if $f(v) \neq v$, then it must be the other.

EDIT: The preimage of $w$ is, $f(w)$ as $f(f(w)) = w$. Moreover, in the second problem, you need to show that if $f(v) \neq v$, then it must be that $f(v) = -v$.

EDIT 2: For the first problem, we need to prove that $V = ker(f) \oplus im(f)$, they clearly intersect in $0$ because if $w \in im(f)$, then $w = f(v)$, but since $f^{2} = f$, we have that $f(w) = f^{2}(v) = f(v) = w$ and if in addition, $w \in ker(f)$, then $f(w) = 0$ and hence $w = 0$. Also, if $w \notin ker(f)$, then $w \in im(f)$ which proves they cover the whole of $V$.

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    wouldn't then be w the image and preimage at the same time? – 2017-02-02
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    but I still don't really know what exactly I have to show – 2017-02-02
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    @user406473 I made an edit. Does that answer you? – 2017-02-02
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    yes I understand it better now, but I still don't understand what exactly I have to proof in the first one. Is it enough to say that because f(f(w))=wf(f(w))=w the claim follows? – 2017-02-02
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    @user406473 i made another edit. better now? – 2017-02-02
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    yes thanks alot! – 2017-02-02
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    Don't hesitate to ask more questions. I am glad to be helpful @user406473 – 2017-02-02
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    for the second one: because f*f=id it means that f(v) can only be v and -v because only these are the only ones which are possible? as id(f(v))=v or -v. So now that we know these are the only options we can conclude that if f(v)≠v it must -v because thats the only possible option left? Can we now say that because f(v) is only v or -v they cover the whole of V? and because it's the same Vector both times only in a different direction they only intersect in 0? – 2017-02-02
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52960/discussion-between-vishal-gupta-and-user406473). – 2017-02-02