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 $E_1E_2=E_2E_1=0$ ?
Prove that a sum of projections is a projection iff they are orthogonal, if the characteristic of the space is not $2$
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0@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
2 Answers
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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1@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
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.