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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. – Freeman Aug 31 '11 at 16:18
@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. – t.b. Aug 31 '11 at 16:28
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. – Freeman Aug 31 '11 at 16:29
@Theo, LHS: Was it not possible to correct the LaTeX edit instead? – Asaf Karagila Aug 31 '11 at 16:33
@Asaf: I'm sorry, but I have never used LaTeX before. – Freeman Aug 31 '11 at 16:35
up vote 4 down vote accepted

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=0 – Freeman Aug 31 '11 at 16:46
No problem; I have been kind of slow myself; If we assume $E_iE_j=0$, and we have $1+1=0$, then $(E+E)^2$=$E^2+EE+EE+E^2$=$E^2+0+0+E^2=2E^2=0$ – gary Aug 31 '11 at 17:13
Yes, I think that works; let me double-check. – gary Aug 31 '11 at 17:25
@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 $E_2E_1 = 0$. – Daniel Fischer May 1 at 15:28
@Arcane1729 Also look at this question, 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 $2X = 0 \implies X = 0$). – Daniel Fischer May 1 at 15:44

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).

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