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Theorem: Given two linear mappings $f,g: V \rightarrow V$ with

  • $f\circ g = g\circ f = 0$
  • $f+g=\operatorname{id}_V$
  • $f\circ f = f$
  • $g\circ g=g$

Then we have $$V=\operatorname{im}(\,f)\oplus \operatorname{im}(\,g)$$


I think $f$ and $g$ then are some kind of projections, where $f$ sets some coordinates to zero and $g$ sets exactly the remaining components two zero. It is clear to me, that the statement then is correct but how to show it formally?

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Hint: In addition to $im(f), im(g)$, also think about $ker(f), ker(g)$. – vadim123 Dec 12 '13 at 18:14
up vote 3 down vote accepted
  1. Since $\operatorname{id}_V = f+g$, for any $v \in V$, $v = \operatorname{id}_V(v) = (f+g)(v) = f(v) + g(v)$. What does this imply about $\operatorname{im} f + \operatorname{im} g \subseteq V$?

  2. Let $v \in \operatorname{im} f \cap \operatorname{im} g$, so that $f(x) = v = g(y)$ for some $x$, $y \in V$. In light of your conditions on $f$ and $g$, what do you learn if you apply $f$ to both sides of the equation $f(x) = g(y)$? What do you learn if you apply $g$ instead?

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Then we have $im(g)+im(f)=V$, because of (2) we have $im(f)\cap im(g)= \lbrace 0_V \rbrace$, thus $im(g)\oplus im(f)=V$ ? – user127.0.0.1 Dec 12 '13 at 18:33
Oh, i see the exactly this in the other answer.. tyvm :) – user127.0.0.1 Dec 12 '13 at 18:35

Choose $ v\in V $. Since $ f+g=id_V $, we have $ f (v)+g (v)=v $. So $ V=im (f )+im (g) $. In order to prove that this is a direct sum, we need to prove that $ im (f)\cap im(g)=(0)$. Choose $y $ in the intersection. Then $ y=f (a)=g (b) $ for some $ a, b\in V $. Then $ f\circ f (a)= f\circ g (b) $. But $ f\circ g=0 $ and $ f\circ f=f $. So $ f (a) =0 $. Hence, $ y= 0 $. Therefore as required the intersection is (0) and $ V=im (f)\oplus im (g) $.

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Hint:$V=kerf\oplus imf$ because $f^2=f$. What is the relation of $kerf$ and $img$?

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