Suppose that $A$ is a Hermitian matrix and that $u,v$ are two vectors. Is there some known function $\kappa(A)$ so that $||u-v|| \leq \kappa(A) |\frac{u^{\\*}Au}{u^{\\*}u}-\frac{v^{\\*}Av}{v^{\\*}v}|$?


UPDATE: Andrew T. Baker has shown that the answer is "no" in general (to take an even simpler counter-example, take $A=I$) - so let's add the **assumption** that $u$ is a simple eigenvector of $A$.



'Condition number' for Rayleigh-Ritz quotient
I think the answer for a general Hermitian matrix $ A $ is no.


Let $ u, v $ be distinct eigenvectors of $ A $ with the same eigenvalue $ \lambda $ and normalized so that $ u^\ast u = v^\ast v = 1 $. Then $ \| u - v \| > 0 $ and 
\begin{equation\*}
 | u^\ast Au - v^\ast A v | = | \lambda u^\ast u - \lambda v^\ast v | = | \lambda - \lambda | = 0.
\end{equation\*}


A concrete counterexample is
\begin{equation\*}
 A = \left( \begin{array}{ccc} 2& & \\\\ & 2& \\\\ & & 1 \end{array} \right), u = \left( \begin{array}{c} 1 \\\\ 0 \\\\ 0 \end{array} \right), v = \left( \begin{array}{c} 0 \\\\ 1 \\\\ 0 \end{array} \right)
\end{equation\*}


ADDED: if we assume all eigenvalues are distinct, then the above argument leads to some kind of bound. Again, $u,v$ are eigenvectors of $ A $ with corresponding eigenvalues $ \lambda\_1, \lambda\_2 $ and normalized with $ u^\ast u = v^\ast v = 1 $. Then
\begin{equation\*}
 \| u - v \| \leq 2
\end{equation\*}
and
\begin{equation\*}
 | u^\ast Au - v^\ast A v | = | \lambda\_1 - \lambda\_2 | 
\end{equation\*}


so that if we choose
\begin{equation\*}
 \kappa(A) \geq \min\_{\lambda\_i, \lambda\_j} \frac{2}{|\lambda\_i - \lambda\_j|}
\end{equation\*}
we can guarantee your inequality will hold.


