sum of projections equivalence over vector space

Question

Let $V$ be a vector space, let $p, q$ projections over $V$. I am trying to prove the statement which says that if $p+q$ is a projection, then $p\circ q=q\circ p=0$.

I could only show that $p\circ q + q\circ p=0$, but could not show the full statement. Any hints or clues as to how to show it? Is it really necesarry to have $V$ a vector space? Or is it also enough to have a module over a ring?

Answer 1

[Solution below]

A hint: take any element in the image of $p \circ q$, i.e. $w = p(q(v))$, and show, by using the fact that $p \circ q = - q \circ p$ and $p \circ p = p$ that $w = -w$. This implies $w = 0$, so $p\circ q$ must be the zero map.

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SOLUTION:

We already have $p\circ q + q \circ p = 0$. Take some $v \in V$ and set $w = (p\circ q)(v)$. By the previous results, we have $w = -(q \circ p)(v)$. Note that projections are linear, so we have $w = p(q(v)) = q(-p(v))$. Hence, $w$ lies in the image of both $p$ and $q$.

Now, for any projection $p$, we have $p \circ p = p$. So, because $w \in \text{Im}(p)$ and $w\in\text{Im}(q)$, we have $p(w) = q(w) = w$. Now we have $$ q(p(w)) = q(w) = w $$ $$ q(p(w)) = -p(q(w)) = -p(w) = -w $$ Hence we have $w= -w$ or $w = 0$. Since $w$ is any arbitrary element in the image of $p \circ q$, we conclude $p \circ q = 0$. Similarly $q \circ p = 0$.

Answer 2

Claim: Let $R$ be a ring and suppose that $p,q$ and $e=p+q$ are idempotents in $R$. Then if $2\nmid \mathrm{char}(R)$, $p$ and $q$ are orthogonal, that is $pq=qp=0$.

Proof: Since $e$ is an idempotent, $e^2=p^2+pq+qp+q^2 = e$, and hence as $p$ and $q$ are idempotents, it follows $pq+qp=0$, or $pq=-qp$.

Now in any associative algebra, if $xy=-yx$, it follows immediately by induction that $xy^n =(-1)^ny^nx$. In particular, $pq^2 = q^2p$, and thus since $q$ is an idempotent, $pq=qp$ and thus $pq+qp=0$ implies $2pq=0$. hence provided $2$ is invertible in $R$, we have $pq=qp=0$ as required.

Note: If $\mathrm{char}(R)$ is even then the claim is false: for example, if $\mathrm{char}(R)=2$ then if $p$ and $q$ are idempotents so is $p+q$ provided that $pq=qp$, i.e. provided $p$ and $q$ commute.