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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 $P_1P_2 = 0$?

Rodrigo de Azevedo
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Hasek
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    Does this answer your question? – twosigma Jun 08 '20 at 01:42
  • @twosigma, yes, it does. – Hasek Jun 08 '20 at 01:49
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    The proof therein is complicated. There is an easier way, by simply computing $(p_1 p_2)^2=-p_1p_2$. Now show that $-1$ can not be an eigenvalue of $p_1p_2$ to conclude. – jijijojo Jun 08 '20 at 01:52
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    Is the following okay? Clearly, since $P_1P_2=-P_2P_1$, pre-multiplying by $P_1$ and post-multiplying by $P_2$ yields: $P_1P_2=-(P_1P_2)^2 \tag {1}$ Similarly, pre-multiplying by $P_2$ and post-multiplying by $P_1$ yields: $(P_2P_1)^2=-P_2P_1 \implies (-P_1P_2)^2=P_1P_2 \implies (P_1P_2)^2=P_1P_2 \tag{2}$ Using $(1)$ and $(2)$, we have $P_1P_2=0$ – Koro Jun 08 '20 at 02:04
  • @Koro, I think it's correct and if you'll post it as an answer I'll accept it. – Hasek Jun 08 '20 at 02:16
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    note the stated result is False in fields of characteristic two. Consider e.g. $P_1 = I =P_2$ with scalars in $\mathbb F_2$ – user8675309 Jun 08 '20 at 02:25
  • In general, in a ring, if $a^2=a, b^2=b$ and $ab=-ba$, then $$ ab=aab=a(-ba)=(-ab)a=baa=ba. $$ Therefore, if the ring is an algebra over a field of characteristic $\ne2$, one can obtain from $ab=-ba$ and $ab=ba$ that $ab=0$. – user1551 Jun 08 '20 at 07:58

1 Answers1

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

$P_2P_1 = 0. \tag{11}$

Robert Lewis
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