Given the LTI system
$$\dot{x}(t) = Ax(t)+Bu(t),$$
with $A \in \mathbb{R}^{n \times n}, B \in \mathbb{R}^{n \times m}$. Is there a way to say (in a qualitative way) how controllable an eigenvalue of $A$ is compared to all the other eigenvalues? Something like: "$\lambda_2$ is more controllable than $\lambda_1$ and $\lambda_4$ but $\lambda_3 $ is the most controllable (for an $A$ with at least 4 eigenvalues)."
My thoughts: We know $(A,B)$ is controllable iff (using the PBH-test) there is no left eigenvector $w$ of $A$ that is orthogonal to the columns of $B$, i.e.
$$ w^\top A = \lambda w^\top, \quad w^\top B = f^\top \neq 0^\top.$$
Now, does the euclidean norm of these $f$'s tell us something about how controllable their corresponding eigenvalues are compared to the other eigenvalues, assuming that all the left eigenvalues $w$ had the same norm?