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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!

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migrated from mathoverflow.net Apr 28 '14 at 18:46

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$\dot\lambda_i$ 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.

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