1. Linear operator proof

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.

2. There are actually several issues with your proof. The major problem is that you never actually constructed the vector $\displaystyle v$. In fact, not every vector will work. For example, it is possible for the image of $\displaystyle T$ to be a line passing through the origin. If you had chosen $\displaystyle v$ to be a vector on that line, then $\displaystyle \{v,T(v)\}$ would be linearly independent (since $\displaystyle T$ would simply stretch out $\displaystyle v$).

With this in mind, you need to be a little more careful. The problem may be easier to handle if you consider three cases: the nullity (i.e. dimension of the kernel) of $\displaystyle T$ is either 0, 1, or 2. See what you can do with each one.

3. I'm not sure I follow. For instance, if the nullity is 2, then doesn't every vector map to the zero vector, and therefore there is no basis for $\displaystyle V$ of the form $\displaystyle (v, T(v))$?

4. Originally Posted by Pinkk
I'm not sure I follow. For instance, if the nullity is 2, then doesn't every vector map to the zero vector?
Yes, and so it is a scalar transformation contradictory to assumption.

Suppose that $\displaystyle v,v'$ are l.i. vectors for $\displaystyle V$. If there existed $\displaystyle \alpha,\beta\in F$ such that $\displaystyle T(v)=\alpha v$ and $\displaystyle T(v')=\beta v'$ then $\displaystyle T(V)$ containing two l.i. vectors is $\displaystyle 2$-dimensional. Thus, if $\displaystyle \dim\ker T=1$ so that $\displaystyle \dim\text{im}(T)=1$ one may assume without loss of generality that there does not exist $\displaystyle \alpha\in F$ such that $\displaystyle T(v)=\alpha v$ and so $\displaystyle T(v)\notin\text{span}(v)$ and so $\displaystyle \left\{T(v),v\right\}$ are linearly independent and thus must form a basis for $\displaystyle V$.

Now try for when $\displaystyle T$ is invertible.

5. If the nullity of T is 0, then doesn't my proof actually work, granted I say the coordinate vectors are just according to any basis of my choice?

Edit: Nevermind, I run into the same problem again.