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Let $V$ be a finite dimensional vector space over complex field , $P_1,...,P_n$ be projection linear operators i.e. $P_i^2=P_i$ such that $P_1+...+P_n=I$ , then is it true that $P_iP_j=O,\forall i\ne j$ ? I can prove it easily if $n=2$ ; but I don't know how to prove in general . I have tried doing by induction but no success . Please help . Thanks in advance

  • This is a special case of http://math.stackexchange.com/questions/507796/problem-with-sum-of-projections/525709#525709 – Ewan Delanoy Nov 28 '16 at 15:56

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The trace of a projection $P$ is equal to the dimension of the target space $\text{Im } P$.

Denoting $p = \dim V$ you get $$\text{tr}(P_1 + \dots + P_n) = \sum_i \dim E_i = \text{tr} I_p = p$$ where $E_i = \text{Im } P_i$. As $\mathbb C$ is of characteristic $0$, the $E_i$ are in direct sum.

For $x=x_1 + \dots +x_n \in V$ with $x_i \in E_i$, I claim that $$P_i(x_1+\dots +x_n)=x_i$$ For the proof, first notice that $$P_i(x_1+\dots +x_n)=y_i$$ with $y_i \in E_i$ as $\text{Im } P_i =E_i$. Then $$\begin{aligned} 0 &= I_p(x_1 + \dots +x_n)- (x_1 + \dots + x_n )\\ &=(P_1 + \dots + P_n)(x_1 + \dots +x_n)-(x_1 + \dots +x_n)\\ &= (y_1-x_1)+ \dots + (y_n-x_n) \end{aligned}$$ as $\bigoplus_i E_i = V$, we can conclude as desired that $x_i=y_1$ for $1 \le i \le n$

based on that, you finally get $P_i P_j=0$ for all $i \neq j$.

Rab
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  • I don't understand why $P_i(x)=x_i$ and how does the conclusion follow from there –  Nov 28 '16 at 13:32
  • @SaunDev That's because $P_i(x)=P_i(x_1)+P_i(x_2)+\ldots +P_i(x_n)$ and in this sum all the terms are zero except for $P_i(x_i)$ wich equals $x_i$ – Ewan Delanoy Nov 28 '16 at 13:34
  • @EwanDelanoy : and why is all the other term $0$ ? because $x_k \in range P_k \subseteq ker P_i $ ? but that's what we are trying to prove here ... or is it for some other reason ? –  Nov 28 '16 at 14:43
  • @SaunDev see my answer at http://math.stackexchange.com/questions/507796/problem-with-sum-of-projections/525709#525709 – Ewan Delanoy Nov 28 '16 at 15:56
  • @SaunDev I added some details to answer your question. – mathcounterexamples.net Nov 28 '16 at 17:52