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Let $f : V \to V$ be a linear operator on a two-dimensional vector space $V$ over $\mathbb{K}$. Suppose that $f \neq \lambda\cdot \textrm{Id}$. Does there exist a vector $v$ such that $\{v, f(v)\}$ is a basis for $V$? Which is the matrix of $f$ with respect to the previous basis?

I showed the first claim, but which is the associated matrix? I think it is

$$ \left[\begin{matrix}0 & a_{12} \\ 1 & a_{22}\end{matrix}\right] $$ where $a_{22} \neq 0$. Is it correct or there is other to say?

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    Any thoughts? You could at least verify yourself that you basically have to show that there is at least one vector that is NOT an eigenvector. So you have to show: If every vector is an eigenvector, then $f = \lambda \operatorname{id}$. – 2017-02-16
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    What are your thoughts on the problem? Have you tried anything on your own here? – 2017-02-16

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