Assume $V$ is a 3-dimsional vector space. And $$ \varphi(t):V\rightarrow V $$ is nondegenerate linear map, and smooth respect to $t$. The eigenvalues of $\varphi(t)$ are $$ \lambda_1(t)\le \lambda_2(t)\le\lambda_3(t) $$ the eigenvectors are $e_1(t), ...,e_3(t)$. Respect to the basis $e_1(t), ...,e_3(t)$ (we always assume $e_1(t), ...,e_3(t)$ can be the basis of $V$, in fact, I feel the nondegenerate of $\varphi(t)$ can guarantee it), $\varphi(t)$ satisfy $$ \frac{d\varphi(t)}{dt} = 2 \left(\begin{array}{1} \lambda_1^2(t) +\lambda_2(t)\lambda_3(t) &0 &0\\ 0 &\lambda_2^2(t) +\lambda_1(t)\lambda_3(t) &0 \\ 0 &0 &\lambda_3^2(t) +\lambda_1(t)\lambda_2(t) \end{array} \right) \tag{1} $$ Then, how to show $e_i(t) =e_i(0)$ for $i=1,.,3$ ? Namely, $$ \varphi(t)(e_i(0)) = \lambda_i(t)e_i(0) $$
PS: I want to explain how to understand (1). First, $\varphi(t)$ is linear map satisfy $$ \frac{d\varphi(t)}{dt} = 2\Gamma $$ where $\Gamma=\Gamma(t)$ also be linear map. When we choose suitable basis of $V$, the linear map $\Gamma$ can be presented as matrix. For example, choose the unit eigenvectors of $\lambda_1(t)...\lambda_3(t)$ as basis of $V$, then, the matrix of $\Gamma$ is $$ \left(\begin{array}{1} \lambda_1^2(t) +\lambda_2(t)\lambda_3(t) &0 &0\\ 0 &\lambda_2^2(t) +\lambda_1(t)\lambda_3(t) &0 \\ 0 &0 &\lambda_3^2(t) +\lambda_1(t)\lambda_2(t) \end{array} \right) $$ Now, we have $(1)$.