Are the multiplicative one-parameter subgroups of the general linear group (i.e., morphisms $\lambda:\Bbbk^\times\to\mathrm{GL}_n\Bbbk$ of algebraic groups) completely classified? The obvious candidates are the $$\lambda(t) := \mathrm{diag}(t^{\alpha_1},\ldots,t^{\alpha_n})$$ for certain $\alpha=(\alpha_1,\ldots,\alpha_n)\in\mathbb Z^n$. Then, we can also consider conjugates of these, i.e. $$(g.\lambda)(t) = g\lambda(t)g^{-1}.$$ Are these all?
Thoughts: One-parameter subgroups are commutative. Hence, if they consist of diagonalizable elements, they can be diagonalized simultaneously. Thus, if the image of $\lambda$ consists of diagonalizable elements, it is of the above form. However, are there one-parameter subgroups that are not diagonalizable?