Im asked to give a definition of diagonalizable for $f: V \rightarrow V$, given that V does not have to finite dimensional.
I could not find the definition for this in the books I searched, anyway I came up with this:
I thought that if i can pick an arbitrarily large set of linearly independent vectors $\{v_1, v_2,..v_n\}$ such that $f(\vec v_i)=\lambda_i \vec v_i$ with different $\lambda_i$ then $f$ is diagonalizable.
Here is my definition:
There is a basis $B$ of $V$ such that for every $\vec v \in B$, $f(\vec v) = \lambda \vec v$ for some $\lambda$ in the base field.
Arbitrarily large $\{ v_1, v_2, \cdots, v_n \}$ is not enough, since the vector space can have uncountable dimension.