Prove that if $P_1, P_2$ and $P_1+P_2$ are projections, then $P_1P_2 = 0$

Question

Suppose that $P_1, P_2$ and $P_1+P_2$ are projections. Prove that $P_1 P_2 = 0$.

Since $P_1+P_2$ is a projection it should satisfy $(P_1+P_2)^2 = P_1+P_2$, i.e. $$(P_1+P_2)^2 = P_1^2 + P_2P_1 + P_1P_2 + P_2^2 = P_1 + P_2P_1 + P_1P_2 + P_2 = P_1 + P_2$$ so $P_2P_1 + P_1P_2 = 0$ or $P_1P_2 = -P_2P_1$. How can I conclude now that $P_1P_2 = 0$?

Answer 1

Assume that $P_1$ and $P_2$ are elements of some operator algebra over a field $\Bbb F$ with

$\text{char} \; \Bbb F \ne 2. \tag 0$

Given that

$P_1^2 = P_1, \tag 1$

and

$P_2^2 = P_2, \tag 2$

and

$(P_1 + P_2)^2 = P_1 + P_2, \tag 3$

observe that

$P_1 + P_2 = (P_1 + P_2)^2 = P_1^2 + P_1P_2 + P_2P_1 + P_2^2 = P_1 + P_1P_2 + P_2P_1 + P_2, \tag 4$

whence

$P_1P_2 + P_2P_1 = 0; \tag 5$

left multiply this by $P_1$:

$P_1P_2 + P_1P_2P_1 = P_1^2P_2 + P_1P_2P_1 = 0; \tag 6$

right multiply (5) by $P_1$:

$P_1P_2P_1 + P_2P_1 = P_1P_2P_1 + P_2P_1^2 = 0; \tag 7$

it follows that

$P_1P_2 = -P_1P_2P_1 = P_2P_1; \tag 8$

also, from (5),

$P_1P_2 = -P_2P_1; \tag 9$

add (9) and (10):

$2P_1P_2 = P_2P_1 - P_2P_1 = 0; \tag{10}$

then by virtue of (0),

$P_1P_2 = 0, \tag{10}$

$OE\Delta$.

Observing the symmetry 'twixt $P_1$ and $P_2$ it also follows that

$P_2P_1 = 0. \tag{11}$