Derivative of eigenvectors of a symmetric matrix-valued function Given a real symmetric $3\times3$ matrix $\mathsf{A}_{ij}$ and its derivative (w.r.t. some parameter, let's call it time) $\dot{\mathsf{A}}_{ij}$, I want to measure/obtain the rotation (rate and direction) of the eigenvectors (the eigenvectors of a real symmetric matrix form an orthonormal matrix). How can this be done?
Edit Since the eigenvectors of a real symmetric matrix are mutually orthogonal, the change of the eigenvectors can only be an overall rotation. An infinitesimal rotation is uniquely determined by the rate $\boldsymbol{\omega}$ such that $\dot{\boldsymbol{x}}=\boldsymbol{\omega}\times\boldsymbol{x}$ for any vector $\boldsymbol{x}$. My question then becomes how to obtain $\boldsymbol{\omega}$.

In order to simplify the algebra, let me first without loss of generality rotate into a coordinate frame where the eigenvectors are the Cartesian basis,
$$
\boldsymbol{\mathsf{A}} =
\begin{pmatrix}
\lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix},
\quad\text{and}\quad
\dot{\boldsymbol{\mathsf{A}}} =
\begin{pmatrix}
\dot\lambda_1 & a & b \\ a & \dot\lambda_2 & c \\ b & c & \dot\lambda_3 \end{pmatrix}
$$
with some coefficients $a$, $b$, and $c$. The above relation for $\dot{\boldsymbol{\mathsf{A}}}$ follows from loup blanc's $\dot\lambda=\boldsymbol{v}^T\cdot\dot{\boldsymbol{\mathsf{A}}}\cdot\boldsymbol{v}$. We now express the derivatives of the eigenvectors as rotation, i.e. $\dot{\boldsymbol{v}}=\boldsymbol{\omega}\times\boldsymbol{v}$, giving
$$
\dot{\boldsymbol{v}}_1 = 
\begin{pmatrix} 0 \\ \omega_z \\ -\omega_y \end{pmatrix},\quad
\dot{\boldsymbol{v}}_2 = 
\begin{pmatrix} -\omega_z \\ 0 \\ \omega_x \end{pmatrix},\quad
\dot{\boldsymbol{v}}_3 = 
\begin{pmatrix} \omega_y \\ -\omega_x \\ 0 \end{pmatrix}.
$$
Inserting these into loup blanc's relation
$
(\boldsymbol{\mathsf{A}}-\lambda\boldsymbol{\mathsf{1}})\cdot\dot{\boldsymbol{v}} = \dot\lambda \boldsymbol{v} - \dot{\boldsymbol{\mathsf{A}}}\cdot \boldsymbol{v}
$
we see that each eigenvalue determines the components of $\boldsymbol{\omega}$ perpendicular to its eigenvector and that $\dot\lambda$ drops out of the relation. We find
$$
\dot{\boldsymbol{\mathsf{A}}} =
\begin{pmatrix}
\dot\lambda_1 & \omega_z(\lambda_1-\lambda_2) & \omega_y(\lambda_3-\lambda_1) \\ 
\omega_z(\lambda_1-\lambda_2) & \dot\lambda_2 & \omega_x(\lambda_2-\lambda_3) \\ 
\omega_y(\lambda_3-\lambda_1) & \omega_x(\lambda_2-\lambda_3) & \dot\lambda_3 \end{pmatrix}.
$$
Finally, generalizing for a general coordinate frame, we have
$$
\boldsymbol{\omega} =
\frac{\boldsymbol{v}_2^T\cdot\dot{\boldsymbol{\mathsf{A}}}\cdot\boldsymbol{v}_3}{\lambda_2-\lambda_3}\boldsymbol{v}_1 +
\frac{\boldsymbol{v}_3^T\cdot\dot{\boldsymbol{\mathsf{A}}}\cdot\boldsymbol{v}_1}{\lambda_3-\lambda_1}\boldsymbol{v}_2 +
\frac{\boldsymbol{v}_1^T\cdot\dot{\boldsymbol{\mathsf{A}}}\cdot\boldsymbol{v}_2}{\lambda_1-\lambda_2}\boldsymbol{v}_3
.
$$
In particular, as pointed out by loup blanc, the derivatives of the eigenvectors, i.e. $\boldsymbol{\omega}$, are only well-defined if the eigenvalues are distinct. However, unlike what loup blanc has claimed in a comment, the values for $\omega_i$ obtained from different eigenvalues agree (this makes sense, since $\dot{\boldsymbol{\mathsf{A}}}$ contains six independent numbers, the same as the number of unknowns $\dot\lambda_i$ and $\boldsymbol{\omega}$).
