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Let $E_1$ and $E_2$ be projections on $V$, a vector space over $F$. Why is if $\operatorname{char}F\neq2$ then $E_1+E_2$ is a projection iff $E_1E_2=E_2E_1=0$ ?

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    You are quite right, I apologize. I have not encountered char before, nor have been taught about it, so the wording of the question confused me.2011-08-31
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    @Asaf: the LaTeX improvement may have increased readability; however, in view of mt_'s comment, LHS had turned the question from a completely wrong statement into a correct one while the LaTeX edit re-incorporated the first version.2011-08-31
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    I rolled back because I was mid way through correcting it when you edited it, i'm very grateful for the edit, but it was the original incorrect version.2011-08-31
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    LHS:Have you tried just doing a "binomial" with $E_1+E_2$, and then using the identity $E_1E_2=E_2E_1=0$?2011-08-31
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    @gary: it is clear why if E1E2=E2E1=0 the statement is correct, but i'm having issue with why if E1+E2 is a projection then E1E2=E2E1=02011-08-31
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    @Theo, LHS: Was it not possible to correct the LaTeX edit instead?2011-08-31
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    @Asaf: I'm sorry, but I have never used LaTeX before.2011-08-31
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    @LHS: If $E_1+E_2$ is a projection, then we must have $(E_1+E_2)^2=(E_1+E_2)$ then some terms must cancel out for the equality to hold.See my answer for more details.2011-08-31

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Think about it this way: $E_1+E_2$ is a projection if it satisfies: $(E_1+E_2)^2=(E_1+E_2)$

(Use $E_iE_j$ to mean the composition)

1)Assume $E_1E_2=0$

We want to show that $(E_1+E_2)(E_1+E_2)=(E_1+E_2)$ This means that $E_2E_1+E_2E_2+.....=(E_1+E_2)$ Can you see the next step?

For the converse, assume $(E_1+E_2)$ is a projection, then it must satisfy $(E_1+E_2)^2=....$

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    Ok, so I see if E1E2=E2E1=0 as this implies E1E1+E1E2+E2E1+E2E2=E1E1+E2E2, implying (E1+E2)^2=E1+E2 But for the converse you have E1+E2 as a projection.. i'm unsure how you can show E1E2=E2E1=02011-08-31
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    Where do we use the fact 1+1 is not equal to 0?2011-08-31
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    Actually, you're right; it seems all we need is $E_1E_2+E_2E_1=0$. Let me see...2011-08-31
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    Ok, how about this: we have, 1. (E1+E2)^2=E1+E2 => ((E1+E2)^2)E2E1=E1E2E1+E2E1 2. E1E2+E2E1=0 => E1E2E1+E2E1=0 => E1E2E1=-E2E1 Combining 1. and 2. gives, ((E1+E2)^2)E2E1=-E2E1+E2E1=0 ((E1+E2)^2)E2E1=0 Can this only always be true if E2E1=0?2011-08-31
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    For the Char2 part, I think we use that If $E_1=E_2$, then $(E+E)^2=0\neq (E+E)$, is not a projection, i.e., we want to cover the case $E_i=E_j$2011-08-31
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    Sorry.. probably being stupid here, but why is (E+E)^2=0?2011-08-31
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    No problem; I have been kind of slow myself; If we assume $E_iE_j=0$, and we have $1+1=0$, then $(E+E)^2$=$E^2+EE+EE+E^2$=$E^2+0+0+E^2=2E^2=0$2011-08-31
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    Ah ok I see, how about this, we have E1E2+E2E1=0, so we can make E1E2E1+E2E1=0 and E1E2+E1E2E1=0, if we subtract them we get E2E1-E1E2=0. Substituting into the first equation we get the desired result?2011-08-31
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    Yes, I think that works; let me double-check.2011-08-31
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    One last, or we can go to chat: there are many ways of showing $E_1E_2+E_2E_1=0$ implies that $E_1E_2=E_2E_1=0$; you can, e.g., multiply by $E_2$ on the left, to get $E_2E_1E_2+E_2E_1=0$, then $(E_2E_1)(E_2+Id)=0$, and similar rewriting for $E_1E_2=0$2011-08-31
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    Thanks very much! I think this is sorted, and at any rate due to the substantial amount of algebra my tutor has set me I think I've spent enough time on this. You have been very helpful, appreciate it greatly!2011-08-31
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    No problem, LHS, glad to help.2011-08-31
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    @gary I cannot see how $E_2E_1(E_2+Id)=0$ implies $E_2E_1=0...$?2016-05-01
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    @Freeman I would appreciate if you could please explain also because I don't understand the above2016-05-01
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    @Arcane1729 That's where $\operatorname{char} F \neq 2$ enters. We can decompose $V$ as $R \oplus K$, where $R$ is the image, and $K$ the kernel of the projection $E_2$ - that works for all projections and all vector spaces, regardless of characteristic. Now you have a simple description of $E_2 + \operatorname{Id}$ - on $R$, it's multiplication by $2$, and on $K$ it's the identity. When $\operatorname{char} F \neq 2$, it follows that $E_2 + \operatorname{Id}$ is invertible. And then you can cancel the invertible $E_2 + \operatorname{Id}$ to get $E_2E_1 = 0$.2016-05-01
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    @Arcane1729 Also look at [this question](http://math.stackexchange.com/questions/507796/problem-with-sum-of-projections), where the argument is given in what I consider a more elegant way (the question assumes real scalars, but what is needed for the last step is just $2X = 0 \implies X = 0$).2016-05-01
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Easy direction: If $E_1E_2=E_2E_1=0$, then $(E_1+E_2)^2=E_1^2+E_2^2 = E_1+E_2$.

Conversely, suppose $(E_1+E_2)^2= E_1+E_2$. Then $$ E_1E_2+E_2E_1 = 0 \tag{1} $$ By multiplying both sides of (1) by $E_1$ on the left, or by $E_1$ on the right, obtain two equalities: $$E_1 E_2+E_1E_2E_1=0,\qquad E_1E_2E_1+E_2E_1=0$$ By subtracting these, $$E_1E_2-E_2E_1=0. \tag{2}$$ Adding or subtracting (1) and (2), we get $$2E_1E_2=0,\qquad 2E_2E_1=0$$ Since the characteristic is not $2$, the factor $2$ can be cancelled.

(This answer is based on this post by Alex).