Let be a linear operator on a vector space of dimension 2. Assume that is not multiplication by a scalar. Prove that there is a vector such that is a basis of , and describe the matrix of with respect to that basis.
So I think I have a proof, any critique and suggestions would be appreciated.
Let be any nonzero vector of . Let and be coordinate vectors for and respectively. Since is not scalar multiplication, we have that the 2 by 2 matrix with the coordinate vectors of and as columns has a nonzero determinant and therefore is invertible, and so if we call this matrix , then given any , there is a unique solution to the equation , and so is indeed a basis for .
I'm not sure how to find the matrix for according to this basis though. Again, any help would be appreciated, thanks.
Edit: Nevermind, I think my proof is entirely wrong since there is nothing saying that the vectors in the space have two components or not. I'm stuck.