Squaring a linear transformation

starcoast
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Homework Statement


Prove that T^{2} is a linear transformation if T is linear (from R^{3} to R^{3}.


So I understand when a transformation is considered linear, but I don't understand what squaring a transformation does. I don't think it means squaring the result of the transformation but I'm not sure how else to think of it. Or maybe I'm forgetting some convenient property that would make this proof short and sweet. Any help is very much appreciated!
 
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starcoast said:

Homework Statement


Prove that T^{2} is a linear transformation if T is linear (from R^{3} to R^{3}.


So I understand when a transformation is considered linear, but I don't understand what squaring a transformation does. I don't think it means squaring the result of the transformation but I'm not sure how else to think of it. Or maybe I'm forgetting some convenient property that would make this proof short and sweet. Any help is very much appreciated!

T^2(x) = T(T(x)).

RGV
 
Ray Vickson said:
T^2(x) = T(T(x)).

RGV

Thank you! I should have guessed that.
 

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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