Linear Transformation Equalities

jnava
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Homework Statement

I need to show the following equalities, where T is a linear transformation and * is the adjoint.

Null(T*T) = Null(T)
Range(T*T) = Range(T)



The attempt at a solution

I know I have to show that Null(T) is a subset of Null(T*T) and then Null(T*T) is a subset of Null T...

Let v exist in Nullspace of T. Then T(v) = 0
Now let v exist in the Nullspace of T*T. Then T*T(v) = 0

Now i am lost, any help would be great. Thanks
 
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OK, so let v in the nullspace of T. Then T(v)=0. Can you show that T*T(0)=0??

The reverse is not so easy. But take T*T(v)=0. Then <T*T(v),v>=0. Now use the property of the adjoint.
 
Thank you!
 

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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