Consider the open set $U \subset \mathbb R^{d \times d}$ of all $d \times d$ square matrices with pairwise distinct real eigenvalues.
For each $A \in U$ we have the eigenvalues $\lambda_i(A)$'s satisfying
$$\lambda_1(A)<\lambda_2(A)< \cdots < \lambda_d(A).$$
Furthermore, the maps $\lambda_i$ are smooth.
Question: Fix $B \in U$ with unit eigenvectors $u_i$'s. I want to prove that there exist a neighborhood $N$ of $B$ such that for each $A \in N$ there exists unit eigenvectors $v_i(A)$'s satisfying the following properties:
- the maps $A \mapsto v_i(A)$ are smooth,
- $v_i(B) = u_i$.
I believe that the proof uses some trick with the implicit function theorem. Nevertheless, the maps that I have tried so far did not work.
You don't need anything like the implicit function theorem; you can just explicitly construct $v_i$ using linear algebra such that it is obviously smooth. Let us assume without loss of generality that the first coordinate of $u_i$ is positive. This implies that $A-\lambda_i(A)I$ has trivial kernel when restricted to the subspace on which the first coordinate is $0$. That is, the $d\times(d-1)$ submatrix of $A-\lambda_i(A)I$ obtained by omitting the first column has full rank, or equivalently one of its $(d-1)\times(d-1)$ minors is nonzero. This is an open condition, so there is some neighborhood $N$ of $A$ such that for all $B\in N$, that same $(d-1)\times(d-1)$ minor of $B-\lambda_i(B)I$ is nonzero, so that the eigenvectors of $B$ with eigenvalue $\lambda_i(B)$ have nonzero first coordinate.
But now we can solve for such an eigenvector explicitly. Namely, using the inverse of our invertible $(d-1)\times(d-1)$ submatrix of $B-\lambda_i(B)I$, we can find the unique vector of the form $(1,x_2,\dots,x_d)$ which is in the kernel of $B-\lambda_i(B)I$ (just apply the inverse to the value of $B-\lambda_i(B)I$ on $(1,0,\dots,0)$ to find what $x_2,\dots,x_d$ need to be). This is a smooth function of $B$, since matrix inversion is smooth. Normalizing our eigenvector to get a unit vector, we see that the function $v_i$ on $N$ that sends $B$ to the unique unit eigenvector with eigenvalue $\lambda_i(B)$ and positive first coordinate is smooth.