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Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$

I think I need to use the fact that if $v \in V$, then $v = (v - \pi(v)) + \pi(v)$

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marked as duplicate by Martin Sleziak, Pedro Tamaroff, Micah, Thomas, Erick Wong Dec 1 '12 at 3:06

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A rather quick method would be using the exact sequence $0\to\ker f\to V\to V/\ker f\to0$. –  akkkk Nov 30 '12 at 20:55
    
Please put your entire equation in the math environment, not just the extra symbols you need. So write \$\pi\circ\pi=\pi\$ instead of \$\pi\$o\$\pi\$=\$\pi\$. –  akkkk Nov 30 '12 at 21:09

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Take $v\in V$. You need to show that it can be expressed as a sum of a vector in the kernel of $\pi$, and a vector in the image of $\pi$, and in exactly one way.

You have actually already shown that $v$ decomposes in this way : $v=\big(v-\pi(v)\big)+\pi(v)$ and $v-\pi(v)\in\mathrm{Ker}(\pi)$, since $\pi\big(v-\pi(v)\big)=\pi(v)-\pi\circ\pi(v)=\pi(v)-\pi(v)=0~$ because $\pi$ is a projection, while $\pi(v)\in\mathrm{Im}(\pi)$. So $v$ is the sum of a vector in the kernel of $\pi$ and one in one its image.

Thus far we know that $V=\mathrm{Im}(\pi)+\mathrm{Ker}(\pi)$, and what remains to be shown is that $\mathrm{Im}(\pi)\oplus\mathrm{Ker}(\pi)$ i.e. $\mathrm{Im}(\pi)\cap\mathrm{Ker}(\pi)=\lbrace 0\rbrace$. So consider $w\in\mathrm{Im}(\pi)\cap\mathrm{Ker}(\pi)$ : by definition, there is a vector $v\in V$ with $w=\pi(v)$, and $\pi(w)=0$, but then $$0=\pi(w)=\pi(\pi(v))=\pi\circ\pi(v)=\pi(v)=w,$$ so $w=0$ and $\mathrm{Im}(\pi)\cap\mathrm{Ker}(\pi)=\lbrace 0\rbrace$, so finally $$V=\mathrm{Im}(\pi)\oplus\mathrm{Ker}(\pi).$$

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Let v be in V. Since π is a projection, π^2 = π. Thus if v is in V, π(v - π(v)) = π(v) - π^2(v) = π(v) - π(v) = 0, i.e., v - π(v) is in Ker(π). So v = (v - π(v)) + π(v) is in Ker(π) + Im(π). Since v is arbitary, this shows that V = Ker(π) + Im(π). To see that the sum is direct, suppose w is an element of the intersection of Ker(π) and Im(π). Then π(w) = 0 and w = π(u) for some u in V. Hence w = π(u) = π^2(u) = π(π(u)) = π(w) = 0. Hence V = Ker(π) ⊕ Im(π).

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"So v = (v - π(v)) + π(v) is in Ker(π) + Im(π)." Please explain. –  akkkk Nov 30 '12 at 20:53
    
What do you mean by $\ker(\pi)+im(\pi)$? Do you mean $\ker(\pi)\oplus im(\pi)$? In that case, equivalence is not clear to me. –  akkkk Nov 30 '12 at 21:07
    
In that case you haven't proven that $\ker(\pi)+\operatorname{im}(\pi)=\ker(\pi)\oplus\operatorname{im}(\pi)$, only $\ker(\pi)\cap\operatorname{im}(\pi)\subset\ker(\pi)+\operatorname{im}(\pi)$ –  akkkk Nov 30 '12 at 21:17
    
Your map $v\mapsto(v-f(v),f(v))$ is a good one. Is it a linear space? What is its dimension? –  akkkk Nov 30 '12 at 21:24

Let's call $\pi$ the projection. So: $\pi^2=\pi \Rightarrow \pi(\pi-Id)=0 \Rightarrow t(t-1)$ is divided by minimim polynomial that nullify $\pi; $ so $ \pi$ is diagonalizzable, so you did it.

$V=W_0 \oplus W_1$ (since Generalized Eigenspace Decomposition Theorem of V)$ = V_0 \oplus V_1$ (since diagonazation)

For further info check: http://www.dm.unipi.it/~benedett/JORDAN.pdf.

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