Critical eigenvalues for smooth family of 2x2 matrices?

Consider the following simple setup: we have a smooth family of symmetric $2\times 2$ matrices $A(t)$, with normalized eigenpairs $(\lambda_1(t),v_1(t))$ and $(\lambda_2(t),v_2(t))$.

Suppose there exists a point $t_0$ such that $$\dot{v}_1(t_0) = v_2(t_0)\\ \dot{v_2}(t_0) = -v_1(t_0).$$

Visually, this seems to imply $\dot{A}(t_0)$ is similar to some multiple of the rotation matrix $\left[ \begin{array}{cc} 0 & 1\\ -1 & 0\end{array}\right]$. This would imply both eigenvalues are critical, i.e. $\dot{\lambda_i}(t_0) = 0$. However, I can't seem to prove this or find a counterexample..

Any insight into whether this should be true would be appreciated!

migrated from mathoverflow.netApr 28 '14 at 18:46

This question came from our site for professional mathematicians.

$\dot\lambda_i$ need not be zero. For example $$v_1 = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix} \quad,\quad v_2 = \begin{bmatrix} -\sin t \\ \cos t \end{bmatrix}$$ then $\dot v_1=v_2$, $\dot v_2=v_1$. They are also normalized. Set $$Q = \begin{bmatrix} v_1 & v_2 \end{bmatrix} \quad,\quad A = QDQ^T$$ where $D$ is diagonal. $A$ is symmetric and analytic in $t$ if $D$ is. But $D$ is otherwise arbitrary.