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Part of a problem in Roman's Advanced Linear Algebra asks to show that if $\rho$ and $\sigma$ are projections over a field not of characteristic $2$, then $\rho-\sigma$ is a projection implies $\rho\sigma=\sigma\rho=\sigma$.

There is a hint which states that $\rho$ is a projection iff $\iota-\rho$ is a projection, so $\rho-\sigma$ is a projection iff $\iota-(\rho-\sigma)=(\iota-\rho)+\sigma$ is a projection.

I can prove the hint, but I don't know how to use it to prove $\rho\sigma=\sigma\rho=\sigma$. I suppose $(\iota-\rho)+\sigma$ is a projection, hence idempotent, and use the equality to determine that $2\sigma=\rho\sigma+\sigma\rho$, but that's all. How can the conclusion be made?

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If you multiply your equation, $2\sigma = \rho\sigma +\sigma \rho$, on the left and right by $\sigma$, then you get $2\sigma = 2\sigma\rho\sigma$. Because the characteristic is not $2$, this implies that $\sigma = \sigma\rho\sigma$.

Now if you go back to the equation $2\sigma = \rho\sigma +\sigma \rho$ and only multiply each side of the equation by $\sigma$ on the left this time, what happens?

(Incidentally, I don't see the point of the hint. You get the same equation from the assumption that $\rho-\sigma = (\rho-\sigma)^2$.)


Added: The fact that the characteristic is not $2$ was used in the argument above, but that in itself doesn't tell us what happens in characteristic $2$. The implication is false in that case. For example, on the vector space $\mathbb Z_2\times\mathbb Z_2$ over $\mathbb Z_2$, define $\rho(x,y)=(x,0)$ and $\sigma(x,y)=(0,y)$. Then $\rho-\sigma =\rho+\sigma =I$ is a projection, but $\rho\sigma=\sigma\rho=0\neq \sigma$.

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    Thanks again, Jonas. I get it now.2012-12-17