Proving Basis of Av with Invertible A Matrix

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If A is an invertible matrix and vectors (v1,v2,...,vn) is a basis for Rn, prove that (Av1,Av2,...,Avn) is also a basis for Rn.
 
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Hi Chris! :wink:

Show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
why don't threads like this get moved to homework?
 
Just lazy mentors!
 
I need to prove that (Av1,Av2,...,Avn) spans and that it is linearly independent but this proof is so confusing to me that i don't even know where to start doing that.
 
You need to show that b* A(v_1) + ... b_n A(v_n) = 0 implies b_1 ... b_n equals zero, right? Well, you know since A is invertible, what is it's kernal?
 

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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