This arose in trying to understand the details of a proof of Hadamard's first variational formula in Terence Tao's "Topics in Random Matrix Theory".
Suppose $A$ is an $n \times n$ Hermitian matrix with simple spectrum $\lambda_1 > ... > \lambda_n$. Suppose that $u_1, ... , u_n$ are corresponding eigenvectors with $||{u_i}||^2 = 1$. Let $A = A(t)$ be a smoothly varying curve in the space of Hermitian matrices with simple spectrum.
It can be shown that for small enough $t$ that $\lambda_i(t) = \lambda_i(A(t)$ vary smoothly, and the we can select $u_i(t) = u_i(A(t))$ that vary smoothly as well. Note that the $u_i$ cannot be chosen canonically since eigenvectors of unit norm associated to an eigenvalue are not unique but some choice of $u_i$ exists.
We therefore have the equations $$ Au_i = \lambda_i u_i, \quad u_i^{*}u_i = 1 $$
Differentiating the second equation yields $$\dot{u}_i^* u_i + u_i^*\dot{u_i} = 0$$
It is then claimed that this equation simplifies to $\dot{u_i}^* u_i = 0$ implying that $\dot{u_i}$ is orthogonal to $u_i$. If we were dealing with the real inner product, I would see why this holds. However, all I can conclude is that the real part of $\dot{u_i}^* u_i$ vanishes. So my question is, why must $\dot{u_i}$ be orthogonal to $u_i$?
It isn’t that $\dot{u}_i$ is necessarily orthogonal to $u_i$, but that $\dot{u}_i$ can be chosen to be orthogonal to $u_i$. From $\dot{u}_i^\ast u_i+u_i^\ast\dot{u}_i=0$, we see that $\kappa_i(t)=u_i^\ast\dot{u}_i$ is purely imaginary. Therefore $v_i=e^{-\int_0^t\kappa_i(t)dt}u_i$ is unit eigenvector that varies smoothly with $t$ and $v_i^\ast\dot{v}_i=0$.
E.g. consider the rank-one orthogonal projector $A=u_1u_1^\ast$, where $u_1=\frac{1}{\sqrt{2}}\pmatrix{1\\ e^{it}}$. While $u_1$ is not orthogonal to $\dot{u}_1$, if we modify it to $v_1=e^{-it/2}u_1=\frac{1}{\sqrt{2}}\pmatrix{e^{-it/2}\\ e^{it/2}}$, we have $v_1^\ast\dot{v}_1=0$.