Given a square function valued matrix $A(\boldsymbol{t})$ whose entries are complex continuous functions of a vector of real parameters
$\boldsymbol{t}$, if $\lambda_0$ is a simple eigenvalue at $\boldsymbol{t} = \boldsymbol{t}_0$ with a corresponding unit eigenvector $\boldsymbol{x}_0$, then for all $\boldsymbol{t}$ near $\boldsymbol{t}_0$ there is a corresponding eigenvalue and unique (normalized) eigenvector that are continuous in $\boldsymbol{t}$. I am looking for a proof or a counter example. I am especially interested in the case when $\boldsymbol{t}$ is a two dimensional vector.
Let $\boldsymbol{\Phi}$ be an $n \times n$ integer matrix with entries $\phi_{pq}\in\mathbb{Z}$. Moreover, let $\phi_{pp}=0$ (diagonal), $\phi_{pq} \neq 0$ (non-diagonal),
$\phi_{pq} > 0;~\forall q>p$, $\phi_{pq} = -\phi_{qp}$ and gcd($\{\phi_{pq}\}_{q>p}$) $=1$. Let $\mathcal{X}$ denote the set consisting of all possible
$k \times k$ principle sub-matrices of $\boldsymbol{\Phi}$, with $2 \leq k \leq n$. Define $|\boldsymbol{A}|_{d+1} = \sum_{p=1}^{k-1} a_{pp+1}$ for any
$k \times k$ matrix $\boldsymbol{A}$. Furthermore, the matrix $\boldsymbol{\Phi}$ must also satisfy the constraint that $\forall \boldsymbol{X} \in\mathcal{X};|\boldsymbol{X}|_{d+1}=x_{1k}$, where $k$ is the dimension of $\boldsymbol{X}$.
Below I give the matrix that I am interested in. It is a rank 2 matrix most of the time. Its largest eigenvalue is always simple. I have an analytic expression for its largest eigenvalue. I need to know whether its normalized eigenvector is continuous?
Let $\boldsymbol{\mathcal{R}}(\boldsymbol{b})$ be an $n \times n$ Hermitian matrix (of continuous Hermitian functions) with entries
\begin{equation}
V_{pq}^{\mathcal{R}}(\boldsymbol{b}) = r_{pq}(\boldsymbol{b}) = A_1 + A_2e^{-2\pi i \phi_{pq}\boldsymbol{b}\cdot\boldsymbol{s}_0};A_1,A_2\in\mathbb{R},
\end{equation}
that is
\begin{equation}
V_{pq}^{\mathcal{R}}(u,v) = r_{pq}(u,v) = A_1 + A_2e^{-2\pi i \phi_{pq}(ul_0 + vm_0)}; u,v,l_0,m_0\in\mathbb{R},
\end{equation}
where $\phi_{pq}$ is the element found in the $p$-th row and $q$-th column of $\boldsymbol{\Phi}$, $\boldsymbol{b}=\left\langle u,v \right\rangle$ and $\boldsymbol{s}_0=\left\langle l_0,m_0 \right\rangle$. Furthermore let, $A_1=1,0<A_2<1$, $l_0 \neq 0, m_0 \neq 0$.
Its largest eigenvalue is
$\lambda = \frac{n(A_1+A_2)}{2} + h$,
where $h=\frac{1}{2}\sqrt{[n^2-4 {n \choose 2}][A_1+A_2]^2+4\sum_{p<q}(A_1^2+A_2^2+2A_1A_2\cos(2\pi\phi_{pq}\boldsymbol{b}\cdot\boldsymbol{s}_0))}$
Best Answer
It is a classical result that you can always manage for the eigenvalues to be continuous, even when they are multiple. For hyperbolic matrices (i.e. with real eigenvalues) the result of Bronshtein ensures Lipschitz continuity for the eigenvalues.
For your question in your framework the eigenvalue is smooth: with $p(X,t)$ be the characteristic polynomial, you have at a simple root $$ p(\lambda_0,t_0)=0,\quad \partial p/\partial X(\lambda_0,t_0)\not=0, $$ so that you can use the implicit function theorem to get near $(\lambda_0,t_0)$ the equivalence $$ p(\lambda,t)=0\Longleftrightarrow \lambda=\alpha(t),\alpha(t_0)=\lambda_0\text{ with a smooth function $\alpha$}. $$
The eigenvectors could be very unstable at multiple roots. Consider the $2\times 2$ matrices $$ A_1=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad A_0=\begin{pmatrix}0&1\\1&0\end{pmatrix} $$ and the smooth matrix $ A(t)=H(-t)e^{-1/t^2}A_0+H(t)e^{-1/t^2}A_1,\quad H=1_{\mathbb R_+}. $ The eigenvectors of $A(t)$ for $t<0$ are $e_1,e_2$ whereas the eigenvectors of $A(t)$ for $t>0$ are $$ e_1\pm e_2, $$ so the (normalized) eigenvectors are discontinuous.