Proving Simple Linear Algebra Statement: A^2 = A implies A is either 0 or I

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Homework Statement


Prove that given a matrix A, and A^2 = A, then A must be either the zero matrix or the identity matrix.

The Attempt at a Solution


By multiplying both sides by A, you can deduce that A = A^2 = A^3 = A^4 ...
From there I think it's obvious that A must be either 0 or I, but I don't know how to start proving it formally.

Thanks very much for your help
-Patrick
 
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Try taking the determinant of A
 
Is there a way to prove it without using the determinant. This exercise is given before determinants are introduced.

Thanks
-Patrick
 
If A^2= A, then A^2- A= A(A- I)= 0.

However, you have to be careful here. With matrices it is NOT true that "if AB= 0 then either A= 0 or B= 0".
 
Is that true? What about A = [1,0;0,0]? This is not zero or I but A^2 = A
 
Ah. I didn't spot that buzzmath. Thank you. The question actually does say either prove or find a counterexample. I was just too sure that it was true.

Thanks
-Patrick
 
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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