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Let $F$ be a sub-field of the complex numbers (i.e. a field of characteristic zero). Let $V$ be a finite dimensional vector space over $F$. Suppose that $E_1, E_2, \dots , E_k$ are projection of $V$ and that $E_1 + E_2 + \dots + E_K = I$. Prove that $E_iE_j = 0$ $\forall$ $i \neq j$.

I have tried this problem for for $n = 2$ as follows:

As $E_1$ and $E_2$ are projection operators, $E_1^2 = E_1$ and $E_2^2 = E_2$. Also since $E_1 + E_2 = I$, multiplying this relation by $E_1$ we get $$ E_1 (E_1 + E_2) = E_1 I = E_1 \Rightarrow E_1^2 + E_1 E_2 = E_1 \Rightarrow E_1 + E_1E_2 = E_1 \Rightarrow E_1E_2 = 0 $$

But what to do for the general case?

Supriyo
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  • Related: http://math.stackexchange.com/q/66977/, http://math.stackexchange.com/q/117702/ – Jonas Meyer Aug 28 '13 at 06:30
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    There is nowhere in the post the word "orthogonal". The assumptions only state that $E_k$'s are projections on $V$. Hence, this post should not be marked as a duplicate to Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow P_{j}P_{i}=0$ –  Aug 29 '19 at 15:41
  • This post should not be marked as a duplicate of the second linked question either: the vector space in this post is over a more general field than the one in Problem with sum of projections. Also, the way this post is written differs tremendously from the linked one. –  Aug 29 '19 at 15:46
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    The first linked question is certainly inapplicable since it is only about orthogonal projections (and all of the answers use that hypothesis). I don't see anything wrong with the second one though. The statement is not exactly the same, but the accepted answer there also answers this question. – Eric Wofsey Aug 29 '19 at 16:38
  • @Eric I removed the one duplicate. I agree for the other. The fields here are more general, but that seems immaterial at least for the main answer on the dupe. – quid Aug 29 '19 at 17:19
  • @Jack I addressed the point regarding the field by a minimal modification of the target. I don't know how the formulation are that different. If needed, we can include the term "projection" in the body of the linked post too, not just the title. In any case different formulations of the same problem are precisely duplicates, though maybe one that should not be deleted, yet still closed. – quid Aug 29 '19 at 18:11
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    Hint: you may find it useful that the trace of a projection is the dimension of its range. – Robert Israel Aug 28 '13 at 07:07
  • Note: first linked and second linked above refer to older dupes. "Second" is the current dupe, "first" got removed (details in the edit history) – quid Aug 29 '19 at 22:00

1 Answers1

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Can you break a vector $x=x_1+\dots+x_k$

$E_1(x)=x_1$ and $E_i(x_j)=0,i\neq j$ and what happen $E_iE_j(x),i\neq j $ ?

Myshkin
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