Is P = P² the definition of a projection operator?

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Ignore this post (it was irrecoverable due to bad LaTeX markup). See the one below.
 
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Here's the question I'm stuck in on my latest problem set:

Suppose P is an operator on V and P^2=P. Prove that V = null P \oplus range P.

I'm pretty sure that P is simply the projection operator (consisting of the identity matrix replacing some 1's with 0's), in which case the conclusion follows easilly. But I've been looking at this one for a while and I can't see how I can prove that P must be the projection operator.

Thanks for your help.
 
I'm pretty sure that P is simply the projection operator (consisting of the identity matrix replacing some 1's with 0's), in which case the conclusion follows easilly.
If I recall correctly, P² = P is the definition of a projection operator!

Not all projections have the form you gave: for example, consider in R² the orthogonal projection onto the line y = x given by [x, y] -> (1/2)[x+y, x+y]
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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