Let $f:\mathbb R^2 \rightarrow \mathbb R^2$ be a linear map and $v=f(e_1), w=f(e_2)$ where $e_1, e_2$ is the standard basis. Suppose $f$ is invertible (otherwise it is guaranteed to be diagonalisable). My question is, is there any "geometric" condition on the vectors $v, w$ which would imply diagonalisability of $f$?
By "geometric", I know I'm not fully rigurous, but intuitively I'm thinking of "calculation-free"/"deducible using only an unmarked straightedge and a compass".
For example, if $A$ is the matrix of $f$ in the standard basis, the condition that $X^2 - Xtr(A) +\det A=0$ has two distinct real roots, or the conditions deduced from this on $v$ and $w$, would not be geometric, since they require calculations using the coordinates of $v$ and $w$.
Edit: maybe a better way to put the "geometrical" part would be this: if we fix the vector $v$ and rotate the vector $w$, and we have a light bulb which lights up when the corresponding transformation is diagonalisable, and switches off when it's not, what pattern do we expect to see in the lighting up of the bulb as a function of the angle of rotation of $w$? Does this pattern fundamentally change with the choice of $v$ and/or the choice of magnitude of $w$?