Matrix of a Transformation Let $\psi$ be a linear operator on $V$, a vector space of dimension two. How do I show, given $\psi$ is not a scalar multiple of the identity, that there is a $v$ such that {$v$,$\psi$(v)$} forms a basis, and further how do I write the transformation as a matrix with respect to that matrix? For the matrix, the first column would be 0,1, but I don't know what the second column would be.
The key observation here seems to be: let $V$ be a vector space over a field $F$ [not necessarily finite-dimensional!], and let $\psi: V \rightarrow V$ be a linear operator on $V$.  Then exactly one of the following occurs:

(i) There is $\lambda \in F$ such that $\psi(v) = \lambda v$ for all $v \in V$.
(ii) There is $v \in V$ such that $\psi(v)$ is not in the span of $v$.

It is clear that the conditions are mutually exclusive.  Seeking a contradiction we suppose that neither (i) nor (ii) holds.  Then there are $\lambda_1 \neq \lambda_2$ in $F$ and $v_1,v_2 \in V$ with $\psi(v_1) = \lambda_1 v_1$ and $\psi(v_2) = \lambda_2 v_2$.  Observe that $v_1$ and $v_2$ must be linearly independent.  Let $v = v_1 + v_2$.  Then $\psi(v) = \psi(v_1) + \psi(v_2) = \lambda_1 v_1 + \lambda_2 v_2$.  Now suppose there is $\lambda_3 \in F$ such that 

$\lambda_1 v_1 + \lambda_2 v_2 = \psi(v) = \lambda_3 v = \lambda_3 v_1 + \lambda_3 v_2$, 

so 

$(\lambda_1 -\lambda_3)v_1 = (\lambda_3 - \lambda_2) v_2$.

Since $v_1$ and $v_2$ are linearly independent, this equation forces $\lambda_1= \lambda_3 = \lambda_2$.  This is a contradiction.

In other words, the only endomorphisms which preserve every one-dimensional subspace are the scalar linear transformations.  The OP's first question is a special case of this.

As for the second question: yes, the first column of the matrix is $(0,1)$.  In fact the second column of the matrix could be anything whatsoever: since $\psi(v)$ is linearly indpendent from $v$, for any $w \in V$ (here I am assuming, as the OP is, that $V$ is two-dimensional) there is a unique linear transformation $v$ such that $v \mapsto \psi(v)$ and $\psi(v) \mapsto w$.
What you need is a v such that
$\psi(v)$ is not a multiple of v, and not zero; (1)
and then you have your basis.

Is there such a v?  There must be a v such that $\psi(v) \ne 0$ (v is not in the nullspace of $\psi$) or else $\psi$ is identically 0.  We will assume this is not the case.

So let v be any vector which is not in the nullspace of $\psi$. There are two cases:

I) v is not an eigenvector of $\psi;$ that is $ \psi(v) \ne \lambda v$.  Then this v  satisfies (1).

II) v is an eigenvector of $\psi$.  Let u be any other non-zero vector which is not a multiple of v.  There must be such a u because this is a 2 dimensional space.  Let w = av + bu where a,b are not zero. 

First, $\psi(w) \ne 0$ because $\psi(v) = a\lambda v \ne 0$.  Next, there must be a choice of a and b such that w is not an eigenvector of $\psi$; if a particular choice of a,b, causes w to be an eigenvector, choose a different b. Then av + bu cannot be a multiple of w; and it cannot be a multiple of v.  Since there are at most two eigenvectors in a 2 dimensional space, w cannot be one.

Then w and $\psi (w)$ are independent and w satisfies (1).