Proving Basis of Av with Invertible A Matrix

  • Thread starter Thread starter Chris Rorres
  • Start date Start date
  • Tags Tags
    Basis Matrix
Chris Rorres
Messages
4
Reaction score
0
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.
 
Physics news on Phys.org
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top