> 
> Let $V$ be a vector space, let $P\_1$, $P\_2$ be linear maps of $V$ into itself. Assume that they satisfy the following conditions.  
> 
>  a) $P\_1 + P\_2 = I $ (Identity mapping)  
> 
>  b) $P\_1 \cdot P\_2 = O$ and $P\_2 \cdot P\_1 = O$  
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>  c) $P\_1 \cdot P\_1 = P\_1$ and $P\_2 \cdot P\_2 = P\_2$  
> 
>  Show that $V$ is equal to the direct sum of images of $P\_1$ ad $P\_2$ respectively.
> 
> 
> 

I think I might be misunderstanding the question. Is it to show for each case that $V$ is direct sum of $P\_1$ ad $P\_2$ or whole set of criteria is required to show it? 

So far I've thought 

For $a)$ Let $v \in V,(P\_1 + P\_2)(v) = I(v) = v \in V $ Also $P\_1 = I - P\_2$ so image of $P\_1$ is null space of $P\_2$ and vice versa. so $P\_1 \oplus P\_2$. 

For $b)$ Since $P\_1 \cdot P\_2 = O$ and $P\_2 \cdot P\_1 = O$, the null space of $P\_1$ is image of $P\_2$ and vice-versa. By rank nullity theorem it follows.

For $c)$ I don't know, I need help here. Also did I make mistake above?