norm of difference of similar matrice

Question

Let $a,b \in C^n$; $A, B\in C^{n\times n}$. If $A$ and $B$ are similar matrices, i.e. there exists nonsingular $S\in C^{n\times n}$ such that $B=S^{-1}AS$, is it possible to proof an inequality in the following form $$ \|B.a-A.b\| \leq Q.\|a-b\| $$ where $Q$ is some expression and the norm is any vector norm?

Answer 1

Take $a = b$. If $Q$ were to exist, we should have that $\|Ba - Aa\| = 0$. Hence,

$$ 0 = Ba-Aa = (B-A)a. $$

Note that this is absurd when $a \not \in \ker (B-A)$. For example, take $a = e_1$, $B = 2I_n, A = I_n$.