Let $V$ be a finite dimensional vector space over a field $F$ of characteristic zero. Let $E_1 , E_2, …,E_k$ be projections of $V$ such that $E_1+E_2+…+E_k=I$. Show that $E_iE_j = 0$ for all $i\neq j$.
Hint: Use the trace function.
Using the hint I got that $\operatorname{trace}(E_i)=\dim(\operatorname{range}(E_i))$ for all $i,\;1\le i\le k$.
Again $I=E_1+E_2+\cdots+E_k \Rightarrow V=\operatorname{range}(E_1)+\operatorname{range}(E_2)+\cdots+\operatorname{range}(E_k)$
Combining both we get $V=\operatorname{range}(E_1)\oplus\operatorname{range}(E_2)\oplus\cdots\oplus\operatorname{range}(E_k)$
After this step I am not being able to progress further. Please help me. Though this problem has already been discussed in a earlier post, there was no hint regarding this approach. So I am posting this problem again. Please do not mark this problem as duplicate. Thank you in advance.
Best Answer
$\text{Range}(E_i E_j) \subset \text{Range}(E_i)$.
Also $E_i E_j = (I - E_1 - \cdots - E_k)E_j = E_j - E_1E_j - \cdots - E_k E_j$, so $\text{Range}(E_i E_j) \subset \text{Range}(E_j) + \text{Range}(E_1) + \cdots + \text{Range}(E_k)$, where this sum misses out $\text{Range}(E_i)$.
Now $\text{Range}(E_i) \cap (\text{Range}(E_j) + \text{Range}(E_1) + \cdots + \text{Range}(E_k) ) = 0$ as the sum is direct, so $\text{Range}(E_i E_j) = 0$.
Hence $E_iE_j = 0$.