Let $\displaystyle T: V \longrightarrow V$ be a linear operator on a vector space of dimension 2. Assume that $\displaystyle T$ is not multiplication by a scalar. Prove that there is a vector $\displaystyle v\in V$ such that $\displaystyle (v, T(v))$ is a basis of $\displaystyle V$, and describe the matrix of $\displaystyle T$ with respect to that basis.

So I think I have a proof, any critique and suggestions would be appreciated.

Let $\displaystyle v$ be any nonzero vector of $\displaystyle V$. Let $\displaystyle (v_{1}, v_{2})^{t}$ and $\displaystyle (w_{1}, w_{2})^{t}$ be coordinate vectors for $\displaystyle v$ and $\displaystyle T(v)$ respectively. Since $\displaystyle T$ is not scalar multiplication, we have that the 2 by 2 matrix with the coordinate vectors of $\displaystyle v$ and $\displaystyle T(v)$ as columns has a nonzero determinant and therefore is invertible, and so if we call this matrix $\displaystyle A$, then given any $\displaystyle b\in V$, there is a unique solution to the equation $\displaystyle AX = b$, and so $\displaystyle (v, T(v))$ is indeed a basis for $\displaystyle V$. $\displaystyle Q.E.D.$

I'm not sure how to find the matrix for $\displaystyle T$ 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.