Suppose that $A = Q\Lambda Q^{-1} \in \mathbb{R}^{n \times n}$ is diagonalizable with possibly complex eigenvalues. Is it possible to find an upperbound $R$ for $\|Q\|_F\|Q^{-1}\|_F < R$ that works for any $A$?
I have tried the following. I can find a lower bound using the submultiplicative property of matrix norm
$$\|Q\|_F\|Q^{-1}\|_F \geq \|Q Q^{-1}\|_F = \|I\|_F = \sqrt{n}.$$
Without loss of generality, assume that the eigenvector matrix contains normalized eigenvectors. In this case, $$\|Q\|^2_F = \sum_{i=1}^n \|Q_i\|^2_2 = n \Rightarrow |Q\|_F = \sqrt{n}$$
If $A$ is symmetric, $Q^{-1} = Q^T$ and $\|Q\|_F\|Q^{-1}\|_F = n $. Therefore, $R$ must be at least $n$. But, how can I find a bound for $\|Q^{-1}\|_F$ for general $A$?